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this is lecture No. 12 of a V222

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electromagnetic field theory

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today will study electro potential

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this will cover Section 4.7 of your tax full

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the learning objectives of today's electric would be

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understand what is electric potential

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and determine the elect

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the potential difference between two points

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what is electric potential

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if we take the help of this analogy

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with respect to this gravitational field

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so in the gravitational field we know that

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moving an object upward against the gravitational field

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increases is gravitational potential energy

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so by default

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we know that due to gravity the force is a downward

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so if we move an object upward

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against this gravitational force

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that in fact we are going to increase

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the potential energy of that object

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because we are acting against him default

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and gravitational feed of the earth

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so an object moving downward within the

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within the gravitational feed

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would lose gravitational potential merging

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so it is the otherwise that if it did it did object

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it is by default moving down

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downward due to the gravity

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or gravitational force of that earth

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then it is going to lose its potential energy

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and if would be the help of some external

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as source extender force

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we are going to move that object

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against this gravitational force

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then we are going to enhance the potential energy

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of that particular object

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so now let's relate this video back to electrostatics

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so what happens that in this particular case

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once you move

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a charge in an electric fees is potential interchanges

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and this is light out and light

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moving a mass object in a adaptation feed

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so in this particular object

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if there's a negative charge created feed

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which is invert and if there's a positive charge

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so by default

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it is going to attract towards the negative chart

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however

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if we are going to move this charge away from this

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negative charge field

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which is opposing field then we are go

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we are going to add add on the uh

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potential energy of this particular field

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so that is the analogy between these two things

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so again repeat that

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it will move this positive charge away from this

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negatively creative charge

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inactive field which is in by default or by default

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an attractive field we are going to

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we're going to work against this inactively charged

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generally electric field and

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we are going to add in the potential energy of this

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our electric potential of this vertical charge Q

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so building on this in order to move

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in order to bring two life charges near each other

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two life charges so by default they dribble each other

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we have to we have to work

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so work must be done in order to separate two charges

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again by default they are going to attract each other

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so by we have to separate these two charges

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so we have to again work for work is again to be done

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so just repeat if there are two like charges

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and they are repelling each other

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and we want to bring them a closer to each other

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then we'll have to do the work

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and if they are they are to oppositely charge opposite

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opposite quality charges

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they are b by default attracting each other

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and we want to sack them off

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then we'll have to apply the work against that

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attraction force so extended force

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F against the electric field

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increases the potential energy

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in these two particular cases

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electrify intensity due to a charge distribution

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can be obtained from Coolum's Law

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in general or because Isla

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when the charge distribution is symmetric

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so this this

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this is the take away from our previous uh

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two collectors

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another way of obtaining electric feed is

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from the electric scaler

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potentially to be defined in this particular section

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and why it is so important

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in a sense

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that this way of finding electric feed is easier

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because it is easier to handle the scaler quantities

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than the vector feed quantity

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and we will see that electric

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scale of potential is a scale of quantity

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so that let's find out what happens

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once we move a charge in the electric field

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of some external charge

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so we want to we want to move a point charge Q

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a distance DL so this is the distance DL

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that we want a point charge Q to be moved

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and the work done is the product of force that we

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apply externally and the distance it is going to cover

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so this is by default the definition of your work

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you know that it is f dot g

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so in this case the travel distances

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the this a differential length segment that is GL

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the external forces

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f that is acting against the external field

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and here the external electric field

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it is it is generated by some source

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you can see that it is the external

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it is the

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it is the electric field against which you are moving

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a charge Q a distance of the L

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so displacement of a point charge

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Q in an electro static field

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which may be generated by another point source

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align source

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or a surface identity or wallet charge identity

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so this we will see in our coming flights as well

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so this is the definition

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that work done is the product of force

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and the distance covered by that particular point

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charge if we move that point charge

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DL distance in the presence of electrostatic field

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due to some other charger

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some other charge distribution and not that negative

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negative sign

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indicates that work is done by an external agent

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this means we are doing the work

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and the force we apply to move the chart

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is against the electric field

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and this is equal and opposite to the force

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which is being exerted by the default

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electrostatic field

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against which we are going to apply the force

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and if we

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want to move the charge in the directional field

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then as a natural

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where as a natural we are not going to do anything

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we don't do any sort of the word and feel does

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why feel does because by default

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either it will be repelling

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or it will be attracting that particular fee

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the total course

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uh total work done are potential energy required

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uh or

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you can see that

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potential energy added on into that particular uh

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charge in moving up

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in moving that charge from point a to point B

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against this electric field

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is found out by intergating

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this particular differential word

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over that entire part that is from a to B

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so that is just uh uh

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from your mathematical point of view

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if you want to find out the complete word

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you just need to uh

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have the integration of that differential element

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over that complete part a to B

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so dividing a

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W by Q is the potential that deposit charge

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and this quantity

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we also define as the potential difference

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between point a and B

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R

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are the work done or equivalently

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it is the work done by the external force

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in moving a unit point

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charge from point a to point B in

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in the process of external electric v

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and mathematically we can define it from here

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we can define it that w

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y q is equal to minus interval of a to B E

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don and b l and it is in the sense

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it is a potential difference

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it is in the universe of wars or in otherwise

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it is Joles Barcula

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so electric potential point of view

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there are some notes that if v a

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B is negative

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if the overall integration is negative

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then there's a loss in potential energy in moving point

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Q from a to B

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and in that case the work is done by defeat

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okay it is a V

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a V narrative

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remember that the work will be done by the field

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there will be no gain in the potential energy

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rather it will be lost in the potential energy

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of that particular charge

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because

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it is moving in the direction of that particular field

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and if we be positive then there's a gain

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important energy of that particular charge

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because in that case we are

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external force is acting upon

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and it is acting against that particular

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electrostatic field

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and it is going to perform the work

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so there is a net gain in the potential energy of that

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queue because we are

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we are we are moving that charge

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using that external force

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against the direction of existing electrostatic fee

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the third thing is VAB is independent of B

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part of integration

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so either you move from here to here

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you follow this part or this part

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the neck

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the neck gaining potential energy will be the same

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in determining this V a

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b

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a is considered by

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conventionally as a mission point by V by

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by B we have taken as a final point

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so this convention

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we are going to follow through this lecture

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so let's take this

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that E is the E which is being generated

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and against which we are going to do the word

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is being generated by by another point chart Q right

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so we know that from kulunzla

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that electrically generated by this particular charge

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if it is present at the origin is Q divide

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therefore perhaps cannot rskr

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and if we want to move

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and if we want to move another point charge

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another point charge from this

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from this point a to B falling this part

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then what is the well done

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so what the well done is or they put change in this uh

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or increase in this potential energy is minus integral

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okay to be edot BL

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and since we are playing in our sperical partner system

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so BL in the sperical partner system is defined at this

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at this and since the electric field which was

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which was generated by this uh

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point charge against which we are doing the work is uh

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is also really outward

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so this start product will be only with this d R

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d R

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hey now we have to integrate this thing

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and if we integrate this thing

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so you can bring this up

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so it is r minus 2 you involve that R rule

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so at a result it will its integration will be 1

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0 r r okay

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now you put in this image to rearrange this minus sign

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and then you will find out that this V a

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B or the potential difference between point B and a

260

00:12:27,066  -->  00:12:29,833
while you're moving this point check Q

261

00:12:30,466  -->  00:12:33,900
while you're moving this point check Q is uh

262

00:12:33,900  -->  00:12:37,100
is a Q divided by 4 by astronaut and uh

263

00:12:37,100  -->  00:12:40,166
multiply by 1 or the distance factor of this

264

00:12:40,166  -->  00:12:40,966
uh

265

00:12:41,666  -->  00:12:45,533
particular the distance factor of this two point

266

00:12:45,533  -->  00:12:48,433
initial and end point points

267

00:12:48,933  -->  00:12:51,466
within which you are moving your point

268

00:12:51,466  -->  00:12:54,300
charge against this existing electrostatic field

269

00:12:54,666  -->  00:12:55,666
so again

270

00:12:55,866  -->  00:12:59,200
these are the position factors for that initial

271

00:12:59,200  -->  00:13:00,800
end and points

272

00:13:03,933  -->  00:13:07,066
so building on the results of previous life

273

00:13:07,500  -->  00:13:13,166
if we choose a reference point with potential of a zero

274

00:13:13,166  -->  00:13:13,966
for example

275

00:13:13,966  -->  00:13:17,900
if we choose a zero reference for potential that is

276

00:13:17,900  -->  00:13:20,800
and that means you could zero at infinity

277

00:13:21,066  -->  00:13:23,533
so we take that v a v

278

00:13:23,533  -->  00:13:26,866
a is equal to zero at position rector

279

00:13:26,866  -->  00:13:28,333
r is equal to infinity

280

00:13:28,333  -->  00:13:31,966
so this means we are bringing a charge

281

00:13:32,366  -->  00:13:36,400
we are bringing that charge from infinity to words

282

00:13:36,400  -->  00:13:38,033
that point b

283

00:13:39,000  -->  00:13:40,066
like this one

284

00:13:41,400  -->  00:13:44,800
and so you can see like this if this is your Q

285

00:13:45,566  -->  00:13:48,033
this is your R B

286

00:13:49,300  -->  00:13:53,900
this is some that you are e and that is very

287

00:13:53,900  -->  00:13:55,466
very far from this

288

00:13:57,966  -->  00:13:59,566
first point charge

289

00:13:59,566  -->  00:14:01,966
which is generated in this electric field

290

00:14:04,400  -->  00:14:06,800
and you're moving another point

291

00:14:06,800  -->  00:14:10,366
chart toward this point b

292

00:14:10,366  -->  00:14:12,100
in the presence of this electrical

293

00:14:12,666  -->  00:14:15,666
so what you can assume that at infinity

294

00:14:15,666  -->  00:14:17,400
the potential is zero

295

00:14:17,400  -->  00:14:20,666
so if you put that uh potential zero over here

296

00:14:20,733  -->  00:14:24,300
you put that uh distance infinity nah

297

00:14:24,333  -->  00:14:28,766
the uh r a to go infinity into this equation

298

00:14:29,866  -->  00:14:33,333
and for that on what you can do

299

00:14:33,333  -->  00:14:38,266
that you just use this mathematical thing

300

00:14:38,266  -->  00:14:42,233
that one divider information equal to 0

301

00:14:42,400  -->  00:14:45,400
so you can simplify the situation into this form

302

00:14:46,400  -->  00:14:49,500
so gently the potential at any point

303

00:14:49,666  -->  00:14:52,866
distance are from a point charge queue

304

00:14:52,966  -->  00:14:54,966
which is placed at origin

305

00:14:55,133  -->  00:14:56,900
and it is generated

306

00:14:56,900  -->  00:14:59,300
some sort of electrostatic field is

307

00:15:00,600  -->  00:15:04,666
v is equal to q d a before by ABS or not into r

308

00:15:05,066  -->  00:15:07,333
and we are assuming

309

00:15:07,333  -->  00:15:10,500
under this Assumption that v is equal to zero

310

00:15:10,500  -->  00:15:11,366
at infinity

311

00:15:11,366  -->  00:15:14,466
is the reference from where you are moving that point

312

00:15:14,600  -->  00:15:17,100
charge towards that point b

313

00:15:17,766  -->  00:15:19,166
the point at so

314

00:15:19,166  -->  00:15:20,900
the potential at any point is

315

00:15:20,900  -->  00:15:24,533
the potential difference between that point right

316

00:15:24,533  -->  00:15:27,766
and a chosen point which is

317

00:15:27,933  -->  00:15:29,300
which can be a reference point

318

00:15:29,300  -->  00:15:32,800
and in this case it is with a potential equal to zero

319

00:15:34,566  -->  00:15:35,633
so if they are

320

00:15:36,500  -->  00:15:39,533
if they are in number of charges or in the

321

00:15:39,533  -->  00:15:41,966
in this case your point charge

322

00:15:41,966  -->  00:15:44,700
which is generating the electrostatic field

323

00:15:44,700  -->  00:15:46,966
it is not located at the origin

324

00:15:46,966  -->  00:15:49,800
it is located at some position Rector Albert

325

00:15:50,133  -->  00:15:51,133
then in order to find

326

00:15:51,133  -->  00:15:54,200
in order to find out the distance

327

00:15:54,200  -->  00:15:58,133
the distance between this source charge

328

00:15:58,133  -->  00:16:00,800
and the destination point b

329

00:16:00,800  -->  00:16:03,900
we just need to use involve this distance formula

330

00:16:03,900  -->  00:16:06,100
and the distance will be equal to B

331

00:16:06,400  -->  00:16:08,300
and the magnitude of this uh

332

00:16:08,500  -->  00:16:12,833
uh distance vector between these two position vectors

333

00:16:13,266  -->  00:16:16,200
so if there are there are a number of charges

334

00:16:16,200  -->  00:16:16,966
you want to Q

335

00:16:16,966  -->  00:16:20,700
and which are located at some position records R1R2

336

00:16:20,700  -->  00:16:21,100
2 r

337

00:16:21,100  -->  00:16:25,200
N and we have to find out the potential at some point

338

00:16:25,200  -->  00:16:27,300
which is located for example

339

00:16:27,300  -->  00:16:31,633
that point is B and that is located at distance Rector

340

00:16:31,700  -->  00:16:33,200
Acquisition Rector R

341

00:16:33,200  -->  00:16:36,400
then we can simply enroll the superposition principle

342

00:16:36,566  -->  00:16:40,066
and we can sum up the potentials which are

343

00:16:41,000  -->  00:16:42,200
which are being generated

344

00:16:42,200  -->  00:16:42,800
which are

345

00:16:42,800  -->  00:16:45,066
which are being generated in the presence of each

346

00:16:45,066  -->  00:16:48,266
of individual point source

347

00:16:48,266  -->  00:16:49,600
and then we can add them up

348

00:16:49,600  -->  00:16:51,266
we can simplify this into

349

00:16:51,266  -->  00:16:53,033
what's mission formation as well

350

00:16:55,066  -->  00:16:58,800
so if we are having uh instead of a point charge

351

00:16:58,800  -->  00:16:59,933
the electr static

352

00:16:59,933  -->  00:17:02,966
fee is being generated by some other type of

353

00:17:02,966  -->  00:17:04,700
charge distribution for example

354

00:17:04,700  -->  00:17:07,566
the continent's charge distributions and those we can

355

00:17:07,566  -->  00:17:10,166
we will know that they can be offline

356

00:17:10,166  -->  00:17:12,500
circus and warning charge distributions

357

00:17:12,533  -->  00:17:14,066
what we need to do do that

358

00:17:14,066  -->  00:17:16,100
we need to simply replace this

359

00:17:16,100  -->  00:17:19,533
cue that we had in our previous slide

360

00:17:19,533  -->  00:17:21,566
with this integral

361

00:17:22,566  -->  00:17:26,500
so it is a integral of uh line charger

362

00:17:26,600  -->  00:17:29,166
line charger uh charge distribution

363

00:17:29,166  -->  00:17:32,533
it is a integral of second charge distribution

364

00:17:32,533  -->  00:17:35,333
it is a integral of volume charge distribution

365

00:17:35,333  -->  00:17:38,400
and integrated over entire volume of taxing

366

00:17:39,000  -->  00:17:40,300
and uh note that here

367

00:17:40,300  -->  00:17:43,066
the primed prime position rectors

368

00:17:43,066  -->  00:17:46,833
they are going to represent the position of the 4.4

369

00:17:46,866  -->  00:17:48,900
source chart distribution

370

00:17:51,133  -->  00:17:54,066
this is another intuition of electric potential

371

00:17:54,066  -->  00:17:57,600
this diagram shows some values of electric potential

372

00:17:57,600  -->  00:18:00,933
at point in the electric field of a positively charged

373

00:18:00,933  -->  00:18:02,200
here so for example

374

00:18:02,200  -->  00:18:04,800
this is a positively charged here it is generating

375

00:18:04,933  -->  00:18:08,466
generating some electrostatic feel and

376

00:18:09,700  -->  00:18:13,533
if you are moving you're moving one column of charge

377

00:18:13,533  -->  00:18:15,766
from point a to point B

378

00:18:15,866  -->  00:18:19,666
since it is another positively charged positive charge

379

00:18:19,666  -->  00:18:20,566
so you are

380

00:18:21,066  -->  00:18:26,166
you're acting against that positively charged

381

00:18:26,166  -->  00:18:28,033
created electrostatic field

382

00:18:28,266  -->  00:18:29,300
so in this case

383

00:18:29,300  -->  00:18:31,666
you are going to gain the potential energy

384

00:18:31,666  -->  00:18:33,733
and if we if somehow the other

385

00:18:33,733  -->  00:18:37,933
we divide these gains in the uh

386

00:18:37,933  -->  00:18:40,966
in this potential energy over some distances

387

00:18:40,966  -->  00:18:42,000
then we just

388

00:18:42,000  -->  00:18:44,466
just for the understanding point point of view

389

00:18:44,466  -->  00:18:48,200
you can see that with a gain of 15 volt

390

00:18:48,266  -->  00:18:51,466
uh 15 uh world or jewels

391

00:18:51,466  -->  00:18:54,566
but uh jewels of potential energy in the States

392

00:18:58,466  -->  00:19:02,100
not for E due to a point charge

393

00:19:02,100  -->  00:19:04,900
we have assumed that um

394

00:19:05,000  -->  00:19:08,100
the zero potential it is at some infinity distance

395

00:19:08,333  -->  00:19:11,066
so in case if it is not an infinity distance

396

00:19:11,066  -->  00:19:13,766
the reference point of the award

397

00:19:13,766  -->  00:19:16,266
it is not at some infinity distance

398

00:19:16,366  -->  00:19:21,000
then obviously it will have some some this

399

00:19:22,333  -->  00:19:23,433
position rector

400

00:19:24,400  -->  00:19:28,400
so in this case what we can do that we can um

401

00:19:28,533  -->  00:19:32,866
we can denote things like this that if another

402

00:19:32,866  -->  00:19:35,966
if any point other than infinity is chosen

403

00:19:35,966  -->  00:19:38,500
as a reference with zero potential

404

00:19:38,700  -->  00:19:39,666
then what we can do

405

00:19:39,666  -->  00:19:44,200
that we can represent that potential

406

00:19:44,600  -->  00:19:47,500
we can represent this other term

407

00:19:47,500  -->  00:19:51,033
in this different thing with some question

408

00:19:51,500  -->  00:19:55,766
right with some question and further

409

00:19:56,966  -->  00:19:59,400
we can generalize this complete thing

410

00:19:59,400  -->  00:20:00,800
for some other points

411

00:20:00,800  -->  00:20:06,200
that the potential at any point are from a point

412

00:20:06,200  -->  00:20:09,400
check Q which is placed that origin is

413

00:20:09,933  -->  00:20:11,300
and and remember that here

414

00:20:11,300  -->  00:20:13,566
this reference point of zero

415

00:20:13,566  -->  00:20:17,000
potential is taken at some distance r e

416

00:20:17,500  -->  00:20:20,900
and why we have taken it as a constant because it is a

417

00:20:21,000  -->  00:20:24,233
it is that fixed it is that fixed position

418

00:20:24,333  -->  00:20:26,733
that's why we have taken it as a constant

419

00:20:26,733  -->  00:20:29,166
eposition factor for this one is fixed

420

00:20:29,166  -->  00:20:30,000
however

421

00:20:30,400  -->  00:20:34,700
since we want to make the evaluations of this potential

422

00:20:35,700  -->  00:20:37,000
towards the destination point

423

00:20:37,000  -->  00:20:38,800
so this destination point

424

00:20:38,800  -->  00:20:41,900
which is at this position director R B

425

00:20:41,900  -->  00:20:43,633
it is varying in this case

426

00:20:43,766  -->  00:20:44,566
however

427

00:20:44,566  -->  00:20:47,566
this part is staying as a constant because we are

428

00:20:47,566  -->  00:20:51,700
we are moving from this point towards the destination

429

00:20:52,900  -->  00:20:56,833
so generally what we can we can write this integration

430

00:20:57,400  -->  00:21:00,266
this integration as this

431

00:21:01,300  -->  00:21:02,733
as a solution as a

432

00:21:02,733  -->  00:21:03,666
as a reverse

433

00:21:03,733  -->  00:21:05,766
as a reverse of this thing as an integration

434

00:21:05,766  -->  00:21:07,466
we can write that this

435

00:21:08,466  -->  00:21:11,666
this thing that it is potential

436

00:21:12,266  -->  00:21:15,500
generally for any electric field digital point source

437

00:21:15,566  -->  00:21:20,566
the potential is equal to minus in Tigorov e dot b L+ t

438

00:21:21,100  -->  00:21:24,566
and here we are assuming that we are only consideringly

439

00:21:24,733  -->  00:21:26,966
this position rector

440

00:21:27,133  -->  00:21:28,066
and this position

441

00:21:28,066  -->  00:21:32,066
rector is going to recover in this constructor

442

00:21:34,566  -->  00:21:36,366
let's solve this example of your book

443

00:21:36,366  -->  00:21:37,466
to further clarify

444

00:21:37,466  -->  00:21:40,233
the concepts of this electric potential

445

00:21:40,266  -->  00:21:43,200
so in this case we are having two point charges

446

00:21:43,466  -->  00:21:47,266
minus 4 microcoolum and 5 microcoolums

447

00:21:47,266  -->  00:21:50,166
and they are located at these lines

448

00:21:50,166  -->  00:21:53,166
at this position factors and what we need to find out

449

00:21:53,166  -->  00:21:56,266
we need to find out the potential at this point

450

00:21:56,266  -->  00:21:59,166
assuming that zero potential at infinity

451

00:22:00,333  -->  00:22:01,533
okay so in this case

452

00:22:01,533  -->  00:22:06,166
it is a superposition problem and Q1 is this

453

00:22:06,166  -->  00:22:07,066
Q2 is this

454

00:22:07,066  -->  00:22:09,500
we just need to plug in the values in this case

455

00:22:09,500  -->  00:22:13,366
and utilize the formula which is for infinity

456

00:22:13,366  -->  00:22:14,466
for infinity

457

00:22:14,466  -->  00:22:17,066
we know that this constant is going to be zero

458

00:22:17,066  -->  00:22:21,900
because we are bringing a charge from infinity

459

00:22:21,900  -->  00:22:26,700
towards this point right towards

460

00:22:26,700  -->  00:22:30,666
toward this point B so in this case it is 1

461

00:22:30,666  -->  00:22:33,033
0 n one

462

00:22:33,200  -->  00:22:36,600
and here is Q1 for example is Q2

463

00:22:36,600  -->  00:22:38,666
they are generated some sort of

464

00:22:40,600  -->  00:22:44,300
electric field so Q1 you can still like attractive one

465

00:22:44,333  -->  00:22:47,066
it is the deposit one like this one

466

00:22:47,466  -->  00:22:48,566
and then you can just

467

00:22:48,566  -->  00:22:51,600
you can just plug in these values of Q1 and Q2

468

00:22:51,666  -->  00:22:54,300
you can find out this uh uh

469

00:22:54,300  -->  00:22:55,600
distance factor between these

470

00:22:55,600  -->  00:23:02,166
uh between this elevation point and this source point

471

00:23:02,900  -->  00:23:04,000
this source point

472

00:23:04,533  -->  00:23:09,400
and just use the magnitude of this thing and what else

473

00:23:09,600  -->  00:23:12,000
and plug in this magnitude over here

474

00:23:12,000  -->  00:23:16,000
and you can find out d potential at this point B

475

00:23:16,566  -->  00:23:19,300
the potential energy or the electric potential

476

00:23:19,333  -->  00:23:21,266
which is attained by uh

477

00:23:21,266  -->  00:23:23,066
once we are moving this point

478

00:23:23,366  -->  00:23:25,066
unit point charge q from

479

00:23:25,066  -->  00:23:28,133
from infinity distance towards this point b

480

00:23:28,133  -->  00:23:30,666
in the presence of this Q and N Q2

481

00:23:32,266  -->  00:23:34,566
let's solve another example of your book

482

00:23:34,733  -->  00:23:36,766
uh and it is a complex one

483

00:23:36,766  -->  00:23:37,866
uh in this case

484

00:23:37,866  -->  00:23:42,100
applying charge 5 nanoculum is located at this point

485

00:23:42,100  -->  00:23:44,533
while another lines

486

00:23:44,533  -->  00:23:47,400
uh uniform uh distribution applying charge

487

00:23:47,500  -->  00:23:49,066
it is uh uh

488

00:23:49,066  -->  00:23:53,166
it is uh located at y equal to 1 and z equal to 1

489

00:23:53,166  -->  00:23:57,166
so it is you can see that it is a x axis

490

00:23:58,800  -->  00:24:04,066
aligned line current distribution

491

00:24:04,766  -->  00:24:06,100
so remember this thing

492

00:24:06,300  -->  00:24:09,133
so the first thing we need to find out the uh

493

00:24:09,133  -->  00:24:10,900
potential at this particular point

494

00:24:10,900  -->  00:24:15,100
if we are moving uh moving 1.1

495

00:24:15,100  -->  00:24:17,366
uh one that uh

496

00:24:17,366  -->  00:24:17,766
uh

497

00:24:17,766  -->  00:24:22,500
point charge from a point which is at zero potential

498

00:24:22,500  -->  00:24:25,433
and this point is located at this zero

499

00:24:25,766  -->  00:24:27,300
so you can assume like this

500

00:24:27,300  -->  00:24:29,700
this is your unit charge right

501

00:24:29,966  -->  00:24:34,266
you're moving this charge to another point a

502

00:24:35,800  -->  00:24:40,233
in the presence of this Q1 and this

503

00:24:41,566  -->  00:24:42,600
what you say

504

00:24:43,200  -->  00:24:47,766
you remember we donate this line chat by this Royale

505

00:24:47,766  -->  00:24:49,533
something like this okay

506

00:24:49,533  -->  00:24:52,366
so the electricity being generated by this thing

507

00:24:52,533  -->  00:24:55,466
and electrocuted engineered by this 9 source

508

00:24:55,466  -->  00:24:57,700
and you are moving from this zero potential

509

00:24:57,700  -->  00:25:00,966
reference point at this origin

510

00:25:01,100  -->  00:25:03,966
and toward this another point r B

511

00:25:03,966  -->  00:25:07,300
you can take it at this point r

512

00:25:07,300  -->  00:25:09,400
B and you can take it as a

513

00:25:11,733  -->  00:25:14,700
your prime you can take in this case okay

514

00:25:14,700  -->  00:25:17,400
so uh since we we can see over here

515

00:25:17,400  -->  00:25:20,066
the potential is being created by uh

516

00:25:20,066  -->  00:25:23,533
two different uh sources the first one is the point

517

00:25:23,533  -->  00:25:25,866
so the second one is the line distribution

518

00:25:25,933  -->  00:25:28,400
and we know that the

519

00:25:28,933  -->  00:25:30,700
you know to find out the potential

520

00:25:30,733  -->  00:25:33,166
we can take the help of this standard equation

521

00:25:33,166  -->  00:25:35,466
that is minus in 2 block e dot DL

522

00:25:35,533  -->  00:25:38,400
and we know that the electric will be tested

523

00:25:38,400  -->  00:25:39,300
you to a point charge

524

00:25:39,300  -->  00:25:43,300
it is Q divider for perhaps or not ask here

525

00:25:43,533  -->  00:25:44,500
and we know we

526

00:25:44,500  -->  00:25:47,566
since it is a sperical corner system problem

527

00:25:47,566  -->  00:25:50,500
so so this dldr yeah

528

00:25:50,500  -->  00:25:53,733
you just uh integrate this thing right

529

00:25:53,733  -->  00:25:54,666
integrate this thing

530

00:25:54,666  -->  00:25:56,800
so minus minus will be cancelled out

531

00:25:56,900  -->  00:25:58,200
so in the integration

532

00:25:58,200  -->  00:26:01,000
you will see that it is minus r 2

533

00:26:01,000  -->  00:26:03,100
you just involve the positive integration

534

00:26:03,100  -->  00:26:06,500
so it will come come out to be A1 or R and minus

535

00:26:06,500  -->  00:26:09,766
minus will be cancelled out so this is your VQ

536

00:26:11,666  -->  00:26:14,166
right but is why this Constance coming out

537

00:26:14,166  -->  00:26:16,466
because at reference point of zero voltage

538

00:26:16,500  -->  00:26:19,766
it is not at some infinity distance

539

00:26:20,800  -->  00:26:22,333
okay for the second case

540

00:26:22,333  -->  00:26:23,100
we know that

541

00:26:23,100  -->  00:26:27,566
electrically generated by an infinite line charge

542

00:26:27,566  -->  00:26:31,166
it is row L divided by 2 by ABS

543

00:26:31,166  -->  00:26:35,366
no not row and it is in this slantrical partner system

544

00:26:35,366  -->  00:26:37,300
remember here and here

545

00:26:37,300  -->  00:26:41,200
if we integrate this one with this back to this row

546

00:26:41,200  -->  00:26:44,600
we know that integral of this one of our row is

547

00:26:45,533  -->  00:26:49,400
which looks like the row is Eleanor Row

548

00:26:50,100  -->  00:26:56,300
so recall your integration concepts over here and okay

549

00:26:56,300  -->  00:26:57,500
so that's it

550

00:26:57,500  -->  00:27:00,600
and here C2 is another integration constant

551

00:27:00,766  -->  00:27:02,366
and we can combine the C 1

552

00:27:02,366  -->  00:27:05,000
c two in the form of another constant that is c

553

00:27:05,666  -->  00:27:08,333
and this is our equation that we will utilize

554

00:27:08,333  -->  00:27:09,500
in the next slide

555

00:27:11,200  -->  00:27:12,533
okay in the first case

556

00:27:12,533  -->  00:27:16,600
we are moving from this origin point at the zero board

557

00:27:16,600  -->  00:27:21,200
to destination point a and uh uh here

558

00:27:21,200  -->  00:27:23,666
uh you can see that what we are going to do

559

00:27:23,666  -->  00:27:26,733
that we are going to simply take the defense

560

00:27:26,733  -->  00:27:30,366
we are going to imply this equation over here

561

00:27:30,366  -->  00:27:32,900
we're going to imply this equation over here

562

00:27:32,966  -->  00:27:36,300
but we are going to put the uh

563

00:27:36,300  -->  00:27:39,666
this uh perpendicular distance uh

564

00:27:40,000  -->  00:27:42,400
uh evaluations and this uh

565

00:27:42,400  -->  00:27:45,500
position after evaluation with respect to this

566

00:27:46,800  -->  00:27:49,233
origin and this destination point

567

00:27:50,933  -->  00:27:52,933
so in this case in this particular case

568

00:27:52,933  -->  00:27:55,966
we remember that this row is the perpendical distance

569

00:27:55,966  -->  00:27:59,333
from x y z to this case any point x y

570

00:27:59,333  -->  00:28:00,866
z to the line

571

00:28:00,866  -->  00:28:03,800
which is the y equal to 1 and the equal to 1

572

00:28:03,800  -->  00:28:07,666
this is x X is a parallel point and this case

573

00:28:07,666  -->  00:28:09,333
if we want to find out the rule

574

00:28:09,333  -->  00:28:12,366
what we need to do that is if this is our uh uh

575

00:28:12,533  -->  00:28:13,700
uh

576

00:28:13,700  -->  00:28:17,866
some uh destination point and this is our source point

577

00:28:17,866  -->  00:28:20,033
so we just need to keep this

578

00:28:20,666  -->  00:28:25,500
take 1 coordinate seem as of your evolution point

579

00:28:25,766  -->  00:28:26,700
and we just need

580

00:28:26,700  -->  00:28:30,200
then need to take the distance between

581

00:28:30,200  -->  00:28:32,500
distance factor between these two points only

582

00:28:32,966  -->  00:28:36,066
so if we generalize this equation

583

00:28:36,800  -->  00:28:37,733
in order to find out

584

00:28:37,733  -->  00:28:39,533
the distance factor between two points

585

00:28:39,533  -->  00:28:41,800
is perpendicular to ax thus

586

00:28:41,900  -->  00:28:43,100
this will be eliminated

587

00:28:43,100  -->  00:28:46,266
because this is a x axis parallel line

588

00:28:46,300  -->  00:28:49,766
so as we remembered we need to keep this coordinate

589

00:28:49,766  -->  00:28:53,166
same as the coordinate of this revolution point

590

00:28:53,700  -->  00:28:56,300
to find out the perpendical distance between this line

591

00:28:56,300  -->  00:28:58,000
and the evolution point now

592

00:28:58,000  -->  00:29:01,100
simply this will be a standardized equation

593

00:29:01,100  -->  00:29:02,200
for this particular case

594

00:29:02,200  -->  00:29:04,700
only for this where distance factor

595

00:29:05,533  -->  00:29:07,800
let's find out this row knot and row a

596

00:29:08,166  -->  00:29:10,100
with respect to this line

597

00:29:10,166  -->  00:29:14,300
and this R knot and r a with respect to this source

598

00:29:14,300  -->  00:29:15,233
point source

599

00:29:16,466  -->  00:29:19,600
so uh in the first case your uh

600

00:29:19,966  -->  00:29:22,233
in the first case that is the doughnut

601

00:29:22,466  -->  00:29:25,000
your knot is your uh uh

602

00:29:25,000  -->  00:29:29,200
your uh your or initiating initialization point

603

00:29:29,200  -->  00:29:30,966
remember it is uh your uh

604

00:29:30,966  -->  00:29:32,966
you can say that your initial point

605

00:29:33,466  -->  00:29:33,900
and here

606

00:29:33,900  -->  00:29:37,233
you see that you're keeping this x coordinates same

607

00:29:37,733  -->  00:29:37,933
and

608

00:29:37,933  -->  00:29:40,166
to find out the potential distance between the line

609

00:29:40,166  -->  00:29:43,900
and this point of origin

610

00:29:44,333  -->  00:29:46,400
and for the second case at this point a

611

00:29:46,400  -->  00:29:47,666
the destination point

612

00:29:48,166  -->  00:29:49,666
we're keeping this coordinates in

613

00:29:49,666  -->  00:29:52,500
to find out the perpetual distance between this line

614

00:29:52,500  -->  00:29:54,100
source and this point a

615

00:29:55,200  -->  00:29:56,100
okay now that

616

00:29:56,100  -->  00:29:59,566
simply solve the distance factor how to find out

617

00:29:59,700  -->  00:30:01,066
find out using this

618

00:30:02,066  -->  00:30:03,300
so the second case is

619

00:30:03,300  -->  00:30:06,900
you need to find out the position vector

620

00:30:06,900  -->  00:30:09,800
or the distance vector between two points

621

00:30:10,133  -->  00:30:12,700
and the first point is your origin

622

00:30:12,700  -->  00:30:15,166
the second point is your destination point

623

00:30:15,166  -->  00:30:18,000
and the source is your unit point source

624

00:30:18,000  -->  00:30:19,400
that is minus 3 4

625

00:30:19,400  -->  00:30:21,400
0 in this vertical example

626

00:30:21,700  -->  00:30:24,600
so that is your uh 5 nanoculums

627

00:30:24,666  -->  00:30:26,666
uh source point location

628

00:30:29,366  -->  00:30:30,533
no uh

629

00:30:30,533  -->  00:30:30,800
you

630

00:30:30,800  -->  00:30:33,200
you know that we don't have the integration over here

631

00:30:33,200  -->  00:30:34,733
integration constant over here why

632

00:30:34,733  -->  00:30:36,800
because we are having minus over here

633

00:30:36,800  -->  00:30:39,900
so the constant will be subtracted in this case

634

00:30:39,900  -->  00:30:40,900
once we uh

635

00:30:40,900  -->  00:30:44,766
will be eliminated once we subtract the uh both uh

636

00:30:46,100  -->  00:30:49,166
both evaluations using the same formulation

637

00:30:50,300  -->  00:30:55,933
may you put in these values and you can uh

638

00:30:55,933  -->  00:30:59,166
you can simplify this thing by putting this value

639

00:30:59,166  -->  00:31:00,400
so not here

640

00:31:00,400  -->  00:31:05,000
the head we have Eleanor run out divide by row a why

641

00:31:05,000  -->  00:31:05,700
why this one

642

00:31:05,700  -->  00:31:07,900
because if you plug in this value over here

643

00:31:07,900  -->  00:31:09,333
you take the difference of this

644

00:31:09,333  -->  00:31:10,866
we know that Eleanor

645

00:31:11,333  -->  00:31:16,100
something like Eleanor a divided by B is equal to

646

00:31:17,466  -->  00:31:18,300
Eleanor

647

00:31:20,300  -->  00:31:23,700
e minus Eleanor B

648

00:31:24,400  -->  00:31:27,833
this is this is your love written property

649

00:31:27,900  -->  00:31:32,000
so you can uh you can write the subtraction of uh

650

00:31:32,133  -->  00:31:34,900
log written term as a division

651

00:31:34,900  -->  00:31:38,166
as a log rhythmic of division of those two terms

652

00:31:39,566  -->  00:31:40,333
your planning value

653

00:31:40,333  -->  00:31:41,966
you are going to find out this value

654

00:31:43,133  -->  00:31:45,000
so the second case in this case

655

00:31:45,100  -->  00:31:48,366
you are having the reference point of 100 word

656

00:31:48,366  -->  00:31:50,700
and it is a place that this point

657

00:31:50,700  -->  00:31:52,600
no you just need to

658

00:31:52,933  -->  00:31:55,600
you don't just need to plug in this value

659

00:31:55,600  -->  00:31:58,566
you just need to really this valuation point

660

00:31:58,566  -->  00:32:00,666
so in this case this is going to stay same

661

00:32:00,666  -->  00:32:02,300
this is going to stay same

662

00:32:02,300  -->  00:32:04,300
this is the distance factor

663

00:32:04,300  -->  00:32:07,266
between your destination point

664

00:32:07,266  -->  00:32:10,066
and your source point right

665

00:32:10,200  -->  00:32:12,033
and what is your

666

00:32:12,300  -->  00:32:16,266
and here your destination point is this one

667

00:32:16,266  -->  00:32:19,900
your source point is initiating point is this one

668

00:32:20,600  -->  00:32:22,733
and you can just plug in these values to find out

669

00:32:22,733  -->  00:32:24,533
this resultant thing

670

00:32:24,533  -->  00:32:27,766
and only differences instead of zero reference board

671

00:32:27,766  -->  00:32:29,833
you are having under board reference board

672

00:32:31,933  -->  00:32:32,600
the third thing is

673

00:32:32,600  -->  00:32:34,666
we want to find out the difference between this

674

00:32:34,666  -->  00:32:36,700
B and C this is simply V

675

00:32:36,700  -->  00:32:37,766
C minus VB

676

00:32:37,766  -->  00:32:41,566
that we have found out in this Part a and Part B

677

00:32:42,133  -->  00:32:44,266
well here we don't need a potential reference

678

00:32:44,266  -->  00:32:46,933
because it will use a common reference

679

00:32:46,933  -->  00:32:48,200
to find out the potential

680

00:32:48,200  -->  00:32:50,266
what is the VC and VB

681

00:32:50,266  -->  00:32:52,800
then this will be eliminated by default

682

00:32:56,200  -->  00:32:58,800
this is your assignment for your week No. 6

683

00:32:59,133  -->  00:33:02,433
and here again note that you have to solve this

684

00:33:04,266  -->  00:33:06,966
practice exercises from 2nd edition

685

00:33:06,966  -->  00:33:08,533
if you're using the 3rd edition

686

00:33:08,533  -->  00:33:09,533
kindly match those

687

00:33:09,533  -->  00:33:13,100
equate those questions with your class

688

00:33:13,333  -->  00:33:15,266
class medals who are on campus

689

00:33:16,366  -->  00:33:17,533
some here today's lectures

690

00:33:17,533  -->  00:33:20,000
so we understood what is the electric potential

691

00:33:20,000  -->  00:33:22,100
how to find out the electric potential

692

00:33:22,333  -->  00:33:25,466
and how to gain the electric potential

693

00:33:25,600  -->  00:33:27,166
how to uh

694

00:33:27,166  -->  00:33:29,700
increase the potential in the presence of some uh

695

00:33:29,733  -->  00:33:31,166
external electric well

696

00:33:31,266  -->  00:33:35,566
electric fields for that particular uh general charge

697

00:33:35,566  -->  00:33:38,400
and then how to uh

698

00:33:38,400  -->  00:33:40,700
find out the potential difference between two points

699

00:33:40,700  -->  00:33:42,366
in the presence of uh

700

00:33:42,366  -->  00:33:44,933
some external fields and this extended field

701

00:33:44,933  -->  00:33:48,166
they can be due to some point source line sources

702

00:33:48,166  -->  00:33:50,900
the charge distribution are such as

703

00:33:50,900  -->  00:33:52,366
all while in charge distribution

704

00:33:55,000  -->  00:33:57,166
so next time we are going to find out

705

00:33:57,166  -->  00:33:59,566
the relationship between electric and

706

00:33:59,766  -->  00:34:01,100
electric field and the

707

00:34:01,100  -->  00:34:03,300
this electric potential and also

708

00:34:03,300  -->  00:34:07,000
we are going to drive the Maxwell's second equation

709

00:34:07,000  -->  00:34:10,800
and we will also study what is an electric and typo

710

00:34:13,366  -->  00:34:14,466
here I thank you all

711

00:34:14,466  -->  00:34:15,766
if you have any questions

712

00:34:15,766  -->  00:34:19,400
they will be entertained to your emails or online

713

00:34:19,400  -->  00:34:20,600
Eastern Corona session

714

00:34:20,600  -->  00:34:23,233
that will have arranged at the department