Free Physics NEET Video Lectures for Electrostatic Potential and Capacitance

00:00:00 e good morning to all of you we continue with our discussion on electrostatics so today we will start a very important topic in electrostatics that is electrostatic potential energy you have all learned about gravitational field and when you have an object in the gravitational field to move the object you need to exert an extra Force so imagine we have uh we have the ground surface here and I need to move an object from a certain height to another height so there is a gravitational for Force which is trying

00:01:31 to pull the object below and I need to apply a force against the gravitational field to move the object up and so I have to do work to move the object from a certain height to a larger height which means that a work which I'm doing actually gets stored in the form of a potential energy of the object so the potential energy I can define a potential energy which means that I the object has some energy here I move the object from one height to another height I moving the object against the gravitational field and I'm

00:02:05 doing work on the object and so when the external force does work on the system the energy of the system increases and so the potential energy of the object increases here so this is a example in gravitational field which you must have studied earlier on in fact if I leave the object here the object falls down on its own accord the field pulls the object down and the potential energy of the object gets converted to kinetic energy we have a similar situation in electrostatics we where we can define an

00:02:36 electrostatic potential energy so suppose let me assume that there is an electric field pointing down uniform electric field pointing downwards uh let me call this as the x-axis this is some axis which called Z axis so I have a uniform electric field pointing downwards so if I have a charge here say take let me take a charge Q so the charge Q will be acted upon by an electrostatic force depending on the sign of the charge the electrostatic force is either downwards or upwards so to move the charge from this

00:03:17 point to another point I have to apply an external force and in that process the external force does work on the system so let me assume that I have an object at this point so let me call this position as x i and this position as XF I stands for initial F stands for final so my electric field is pointing downward so as you can see from here electric field is some constant times I cap with the minus sign the electric field is pointing in the minus X Direction X direction is pointing upwards so what is the force

00:03:55 electrostatic force on the on the charge it is minus q e I cap that is electrostatic force to move the object to move the charge from XI to XF I need to apply an external Force which is negative of this which is minus F electric which is equal to q e i so I have to apply a force in the upward Direction here to move the to move the charge so that is the force that I have to apply externally and what is the distance I must move the distance I must move is uh so let me call this uh L Vector is XF - x i

00:04:42 into I cap so I start from x i and move to XF and the I'm moving like this from here to here so the length which I'm moving the The Distance by which I'm moving is XF - x i into I cap so the work done by the external force on the charge is equal to F external dot L which is equal to Q E into XF - x i so if I call this height H this is equal to q e H so this is the work done by the external force to move the charge from XI to XF and this work gets stored in the form of potential energy of the

00:05:44 charge please note that this is the work done by the external force on the charge so this equation is true for all kinds of charges so let me try to draw a figure here which which sees what happens with various chares charges so I have the electric field pointing down so suppose my charge was positive and I move from here to here so the work done the work done is equal to q e h and Q is positive so w is greater than zero that means I need to do work on this charge in taking it from this point to this point XI to XF

00:06:30 so if I call the initial energy of the charge as UI and final energy as UF you see here in this case UF is greater than UI the potential energy at this point is more than potential energy at this point so I have to do work on the charge in taking the positive charge from here to here now what will happen if the charge was negative so if the charge was minus Q again x i to XF now in this case the work done is equal to q e h is less than zero because Q is negative Q is negative here so if the initial energy is again

00:07:26 UI and final energy is UF in this case UF is less than UI so I'm moving like this so in moving a positive charge from XI to XF I have increased the potential energy of the charge in moving the charge from XI to XF with a negative charge the potential energy has got decreased what is happening what is the difference between these two cases in this case the positive charge there was an electrostatic force on the positive charge which was downwards and I am moving the charge against the electrostatic force here for the

00:08:08 negative charge the electric field is pointing down and I'm trying to move up the force on the negative charge is upwards and so my movement is along the direction of electrostatic force and in which case actually the field does work on the charge and so the charge comes out to be negative the work done comes out to be negative which means that the final energy of the charge is less than the initial energy so you can do similar plots for example if the electric field was pointing downwards and my charge was going from

00:08:41 here to here this is Xi this is XF if this is a positive charge then I'm moving in the direction of electric field so you can show that UF is less than UI in the other case where e is like this and if I have a negative charge and going from XI to XF if by moving like this the electrostatic force on the charge is upwards and I'm moving against so UF is greater than UI here so depending on the direction of electrostatic force if I have to move the charge against the electrostatic force the work done by the external

00:09:23 agent on the charge is positive and so the potential energy increases if the movement is along the direction of the charge of of the of the electrostatic force the work done is negative and the potential energy decreases this is also consistent with gravitational uh situation where if I have to move against the gravitational field the potential energy increases if I move to along the direction of the gravitational force the potential energy decreases the difference between gravitational and electrostatics is in

00:09:51 gravitation there is only one force of attraction there is no fors of repulsion while in the case of electrostatics you can have either attraction or repulsive forces and so you can have combinations of positive and negative charges moving against the electric field or along the direction of the electric field so I leave it as an exercise to you to draw similar figures for E pointing upwards so you may practice electric field is pointing upwards you find out what are the initial and final energies

00:10:32 whether they increase or decrease depending on the charge plus Q or minus q and whether you are moving upwards or downwards so I leave this as a small exercise to you and that will help you to understand whether the potential energy increases or decreases in the various situations now in the example which I looked at I just took the charge vertically up from one point to another point now it may so happen that my movement may not be vertical so let me consider the following problem this is my X this

00:11:07 is z axis so electric field is again pointing downwards uniform electric field everywhere so my objective is to move from one point to another Point here so let me call the coordinates of this point x i zi and this is XF ZF so I start from x i z i and I have to move to this point so let me call this a and let me call this B so I have to move a charge let me take a positive charge I have to move a charge plus Q from A to B now obviously I can have different paths to take the charge from A to B I

00:11:45 can move for example first horizontally and then go vertically let me call this path one I can go vertically first and then go horizontally let me call this path two or I can go straight straight like this path three so I can either take the charge first horizontally here and then take it up or I can take it first up to the same height and then horizontal or I can take it at an angle from A to B straight let so let us try to calculate what is the work done in moving the charge from A to B along these three paths so let me look

00:12:21 at path one now so path one so let me call this C and let me call this D so path one first I go from a to c and then from C to D so in going from a to c let me calculate the work done so uh the electric field as we as before is minus e i l Vector which I'm moving from a to c is is equal to ZF minus z i into K cap remember that c the coordinates of c are actually x i ZF so I not changed the x coordinate I've only changed the Z coordinate and so this Vector connecting a to c is nothing but ZF minus z i into K cap so

00:13:12 the work done in moving charge from a to c is equal to uh so F external do DL dot sorry L and F external is minus F electrostatics dot L Vector which is equal to plus e i into Q the charge which I'm having so the force is Q * electric field dot ZF minus z i into K cap which is equal to Zer because I do K is equal Z so I am actually in moving from a to c I don't have to do any work which becomes a little obvious because now the electrostatic force is vertical and I moving horizontally so the displacement is

00:14:15 perpendicular the electric field the displacement is perpendicular to the force that I'm applying that I have to apply and so I don't have to do any work in taking the charge from a to c now from C to B I have to calculate the work done and this will already done before calculating the work done to move vertically we have just calculated and that happens to be so work done is equal to Q E into XF - XI so the total work done in moving from A to B via C that means a to c and this C to B total work done is actually equal

00:14:54 to Q E XF minus XI now let me try to calculate for this other path path two which is going from a to d and then from D to B so first is a to d so a to d is essentially going from this point vertically up so the Z coordinate of these two are equal x coordinate has changed so just like before what we did the charge is taken vertically up so from going from a to d the work done is equal to Q E into xf- XI in going from D to B just like in going from a to c I have to move perpendicular to the electrostatic field

00:15:58 and so no work is being done so D to be work done is equal to zero so total work done by the external force in moving from A to B yd is equal to q e xf- x i which is the same as the work done in moving the charge from A to B via C these two work done are equal so now let me calculate what is the work done in going from A to B along path three that path three is now not straight it's at an angle so path three so path three I have essentially I'm moving along an oblique direction from A to B so let me calculate this

00:17:06 Vector L Vector so for path three L Vector is equal to XF - x i into I plus zv minus z i into K cap look at this point this point has coordinates x i z i this point has coordinates XF ZF so this Vector joining x i z i to XF ZF is XF - XI I plus ZF minus z i k cap and the external Force again is equal to minus of a electric force which is equal to q e i so work done is equal to F external do L which is equal to q e not I dot XF - x i i plus ZF minus z i k cap which is nothing but q e xf- x i because i. KC is equal to

00:18:20 Zer which is the same as the work done in taking the charge from A to B via C so what we have seen for the electrostatic case is the work done in going from A to B is the same whether I go from A to B via C or via d or straight from A to B in fact you can take any arbitrary path connecting A and B and you will find that the work done in moving from A to B is exactly the same as before Q E not XF minus XI so what this shows is shows us is that the work done is independent of path followed that's a very very important

00:19:15 concept these such forces are called conservative forces gravitation force is also a conservative Force the work done in moving a mass from one position to another position is independent of the path taken and it is a an example of conservative Force so if you have to take a charge from One initial point to another final point it does not matter what is the path you follow let me take as another example the because in this case what we had taken was an electrostatic force which was uniform now let me take a case

00:20:00 in which the electrostatic field is not uniform and that is a point charge so uh let me take a point charge Q so here is my charge so I want to move from an initial point this distance RI to a final position RF okay that's my so I want to move from a charge from here to here so let me assume the charge is Cap small q so I want to move a charge along the radial direction from point R to a distance RF so let me call this a and let me call this B so I want to move a point charge with uh with magnitude Q from a point a to a

00:20:53 point B so again so what is the electrostatic force is Q * e which is equal to q q by 4 Pi Epsilon 0 r² into R cap that is the electrostatic force on the charge so my external Force which I have to apply is minus of electrostatic force which is equal to minus QQ by 4 Pi epon 0 r² R cap and my Vector which I'm moving is along the radial Direction so you know that the work done if the force is non-uniform integral F external dot DL from the initial Point RI to RF from the initial point to the final

00:21:53 point the dot product of force external force that I'm applying do DL vector and in this case as you can see DL Vector is nothing but Dr Vector which is actually equal to R Dr R cap Dr R because the vector is along the RC cap Direction This Is The Arc cap Direction remember we had discussed in kum's law that is RC cap Direction so the work done now becomes by external force is equal to integral F external do DL from uh r i to RF which is actually equal to now this is uh uh the exter the force is minus

00:22:51 QQ by 4 Pi epon 0 r² um R cap dot d r r cap from r i to RF which is equal to minus QQ by 4 Pi epon 0 integral r i to RF Dr R by r² which is equal to - QQ by 4 Pi epon 0 integral of 1 by r² is - 1 by r r i to RF which is nothing but Q Q by 4 Pi epon 0 into 1 by RF - 1 by RF so this this minus signs cancel off and I get QQ by 4 pepon 0 into 1 by RF minus 1 by R that is the work done by the external force in moving the charge small q from this point A to point B where R is the initial distance and RF

00:23:53 is the final distance now this equation is valid irrespective of the sign of the charges so you can see that for example if the charges were both positive so the charge capital Q is positive and the charge which I'm moving is also positive and I move from here to here so this is a and this is B you see with the Positive the force is repulsive the force is in the in this direction the electrostatic force and I'm moving in the same direction so the uh the initial the final potential energy will be less than the initial

00:24:33 potential energy as you can see here the work done because work done was equal to q q by 4 pepon 0 1 by rfus 1 by r i so here q and Q are both positive RF is greater than r R so work done is is negative so the field is doing work already the the force is repulsive and I'm moving along the direction of the force and so actually the field does work on the system and I have to do negative work similarly you can calculate suppose I had a plus Q here and I had a minus Q charge here in going from A to

00:25:26 B now uh Q * Q becomes negative so work done is equal to so this is uh minus Q capital Q by 4 Pi epon 0 1 by rfus 1 by r i this is negative this is negative sign so this is greater than zero so U is greater than U again this is obvious because this negative charge the electrostatic force is attractive and I have to move the charge against the electrostatic force so I have to do work on the charge in moving from A to B in which case that work which I do get stored as potential energy of the charge so you can work out the other

00:26:13 combinations here to find out what is the work done by the uh the external agent in moving the charge from an initial point to final point you can take different cases where RI is less than RF RF is more than R etc etc and I leave the discussion to you now just like in the earlier case what we will find is so let me take for example this the charge Plus for examp Q and I want to move from this point to another Point here so in the earlier example which we started discussing the two lines the two

00:26:53 initial and final points were along the same radial line now I'm I'm taking two points which are not along the same radi radial line a b like this so I have to move like this now I want to move from A to B so I can do different kind of combinations for example first I can move like this along a circular Arc around q and then move from here to here I can move from here to here and then move along the circular Arc or I can have any combination of position paths for example I can do like this I can move along radial then go

00:27:32 like this then move along radial then go like this then move along radial go like this now for example if you look at this path around this path which is a circular Arc the movement of this path is exactly perpendicular to the electrostatic force at every point at this point electrostatic force is if this is a positive charge electrostatic force is like this and I'm moving perpendicular at this point the electrostatic force is like this and I'm moving perpendicular at this point the El force is like this

00:27:59 I'm moving perpendicular T is this point and then I move in the same direction as the electr static Force so here I don't do any work and here I do the work which I have already calculated as q q by 4 pepon 0 1 by rfus 1 by R similarly if you look at this path I go from the distance R to distance RF along the radial Direction and then I move move along the circular Arc connecting this point to this point along this direction I do the same work as this along the arc I do no work because I'm moving perpendicular to the

00:28:40 electrostatic force similarly you can calculate for any path what you'll find is path independence of work done so any electrostatic force that you have the work done to move a charge from One initial point to a final point is independent of the path and as I mentioned before this is a characteristic of conservative forces so what does this imply for example so let me take a uh let me take a following example so I have a charge and then I am moving I I want to move between path A to B like this and I want to come back

00:29:30 so let me call this uh C1 and C2 so I do a circular path so I start from a go to B and come back to a from another path along another path so let me calculate the total work done so total work done is f external do DL and this as I mentioned before this um circle on the integral sign means it's a closed path so this is equal to A to B along C1 F external. DL plus b to a along C2 F external DL now I know that as we have discussed integral A2 b along C1 F external. DL is equal to integral A to B along C2 the

00:30:34 second curve F external dodl which is actually minus of B2 a along C2 F external DL so the work done in going from here to here is the same as the work done in going from here to here along this path the work done in going from here here is negative of the work done in going from B to a that's what this says so you see that these two are equal and of opposite sign which means that I get the following very important uh expression that integral F external do DL is equal to Z that means in a closed path net workk

00:31:22 done in a closed path is equal to Z and this is again a characteristic of conservative forces the net work done in carrying a charge from A to B back to a through any two different paths any path is actually zero so you may be doing some work in taking a to A to B along C1 and then the field does an equal amount of work in bringing the force bringing the charge from B to a along another curve so the net work done becomes zero so this is again a part of conservative forces and so what we can do is with this we can

00:32:08 Define we can define a potential energy we can now Define a potential energy electrostatic potential energy so let me again take a point charge remember we had calculated the uh so work done by external force is equal to q q by 4 pi0 1 by rfus 1 by r okay so remember in gravitational fields we Define the potential energy with respect to some reference so normally we say the ground has zero potential energy so when I take a object by a certain height I do work on the object and so there is a potential energy stored in the object

00:33:19 similarly we will Define the electrostatic potential energy at any point with respect to a reference potential which is so the reference point is usually chosen reference point is usually chosen at infinity and this is zero of potential energy so I will assume that if I have a positive capital Q charge here if the other charge is at Infinity then there is no energy in the system there's potential energy is zero and then when I bring the second charge close to this first charge I will do work on the

00:33:59 charge and I will change the potential energy so in this equation if I substitute R is equal to infinity and RF is equal r that means this will give me the work done in bringing a charge from Infinity to this point set distance R from here so work done in bringing a charge small q from Infinity to this point at the distance small R will be given by this equation which I will Define as the potential energy here which is potential energy is potential energy at a position R between these two pair of charges is

00:34:36 QQ by 4 pi epsilon0 r please remember because the work done is independent of path followed no matter how I bring the small charge Q from Infinity to this point I will have to do exactly the same amount of work so I can define a potential energy which only depends on this point which is the energy to bring a charge small q from Infinity to this point no matter along which path so this this becomes the potential energy at this point so this is the potential energy of a pair of charges separated by

00:35:15 a small a distance small R so if I have a charge capital Q capital Q and small q separated by distance R the potential energy so this is the potential energy of this pair of charges please remember that if I had calculated the following suppose I say I have a chart small q and I bring capital Q charge into this point at a distance R from this small charge I will get the same potential energy the energy is stored as a in the entire electrostatic system of charges itself and it only depends on the

00:36:16 separation between these two charges through this equation QQ by 40 R and this is the potential energy of this system of charges although I have brought in small q from Infinity to this point I could have brought capital Q from Infinity to this point I could have moved both charges from Infinity together to bring it to a certain distance small R whenever finally they land up with a separation of small R this pair of charges now has a potential energy which is QQ by 4 pi0 R so this is for a pair of charges I can

00:36:54 similarly calculate for more charges so for example suppose my final system has three charges so I have suppose I have to bring I have to bring three charges q1 Q2 and Q3 my question is what is the potential energy of this system of three charges so I do the following um procedure first I have nothing here so I bring charge q1 and place at this point because there is no other charge present I have to do no work in bringing q1 and placing it here here once having placed q1 here I bring Q2 from Infinity to this point so let me

00:37:34 call this distance R12 so what is the potential energy between q1 and Q2 so let me call this u12 is equal to q1 Q2 by 4 Pi epon 0 R12 product of the two charges divided by 4 pi0 into separation between them that's the potential energy of q1 Q2 pair of charges now with these two charges present here I now bring Q3 from infinity and place it at this point so let me call this distance as r13 and this distance as R23 now you see when I bring in Q3 from Infinity I have to do work against the electric field of both q1 and Q2

00:38:24 simultaneously so what will happen is the work done by me to bring in charge Q2 will be essentially so let me let me calculate sorry charge Q3 let me calculate the work done in bringing Q3 to this point so this will be W will be integral F external do DL which is uh from Infinity to this point Infinity to this point so let me call this point c a b c so this is equal to minus of of f electric field. DL from Infinity to C and F electric field is actually the charge both the charges now produce electric field at

00:39:10 this point at as a function of distance and we know that we electrostatic satisfies principle conservation so I have q1 by 4 Pi epon 0 r² R cap plus Q2 let me let me write like this okay so F electrostatic is equal to q1 * E1 electric field because of charge q1 sorry Q3 the force on this charge Q3 is Q3 * E1 plus Q3 * E2 E1 is electric field produced by charge q1 E2 is electric field produced by charge Q2 this is the force of force on the charge Q3 because of electric field of q1 this is the force

00:39:58 on the uh charge Q3 because the electric field produced by Q2 and so this is the total electric field that the charge Q3 has to move against so the work done will be equal to minus Q3 * e1. DL Infinity to C minus Q3 Infinity to C2 now what is this this is the work done in bringing a charge Q3 this is the work done in bringing a charge Q3 because of the electric field produced by q1 there is no dependence on Q2 here and that must be as we as we have discussed before q1 Q3 by 4 Pi epon 0 r13 because E1 is the electric field

00:40:55 produced by q1 and this is the work done in bringing Q3 from Infinity to this point at a distance r13 from here so the work done is exactly the same as this equation except that Q2 gets replaced by Q3 and this distance R12 gets replaced by r13 so I get a for work done which is q1 Q3 by 4 pi0 r13 and this is nothing but the work done in bringing charge Q3 from Infinity to this point because of the electric field generated by Q2 so this must be equal to plus q1 Q2 Q3 by 4 Pi epon 0 R23 so this is the work done in bringing

00:41:40 Q3 to this point so the total work done in assembling this charge distribution is the sum of the work done in bringing charge Q2 to this point and the work done in bringing charge Q3 to this point so the total work done or total potential energy is equal to q1 Q2 by 4 pi0 R12 plus q1 Q3 by 4 pi0 r13 plus Q2 Q3 / 4 pi0 R23 so this is the energy because of q1 Q2 this is the energy because of q1 Q3 and this is energy because of Q2 Q3 that's the total energy of the system of charges and as I mentioned we associate

00:42:45 this potential energy with the entire system of charges I have now brought first I brought q1 then I brought Q2 then I brought Q3 you can calculate and show that this is independent of whether I bring q1 first or Q3 first as long as finally the charge distribution sits in this particular pattern the work done it will be equal to this and so the total potential energy stored in the system is the sum of these three so it's independent of the sequence of bringing of these charges so in general you can

00:43:21 write you can actually generalize this to any number of point charges and you get the total potential energy so having looked at potential energy we now we now introduce a very very important Concept in electrostatics again which is the electrostatic potential so electrostatic potential is the work done by an external force in bringing a unit positive charge work done by an external force in bringing a unit positive charge from Infinity to the point is the electrostatic potential at that point so it it is essentially the work

00:44:58 done in bringing a unit positive charge from Infinity to that point so remember we had discussed if I had a charge capital Q and if I have a u if I bring a small charge Q here at a distance R from here the potential energy was U was Capital q small q by 4 Pi Epson 0 into 1 by rfus 1 by r i so this is the potential energy at this point or change in energy in bringing a charge from R to RF so the potential energy was when I when I put RI is equal to Infinity so potential energy U becomes QQ by 4 pi0 r at this

00:45:43 point so if I take a small charge capital Q as unit charge then I will Define the potential at the point R as V of R is equal to Q by 4 pepon 0 r this is for a point charge the potential of a point charge is essentially given by Q by 4 pi0 R so if you have a positive charge here as you move along this direction R increases and the potential potential keeps on decreasing so this is decreasing like this for a negative charge the potential will be if I move along the same direction for a for a negative charge the

00:46:42 potential will be increasing so depends on the sign of the charges whether the potential increases or decreases uh as you move away from the charge so this is the potential at any point is defined for this is the potential for a point charge so for any general electric field distribution you can define an electrostatic potential as the as the work done in bringing a unit positive charge from Infinity to that point so if you have to for example if I had a positive if I have a point charge Q so if I um if this point is say r i

00:47:18 and another point is RF what is the work done in bring in taking a charge from ra to RF so work done by the external force will be equal to the potential at RF minus potential at RI for to bring to take a unit charge unit positive charge so difference in potential gives me actually the work done in taking a unit positive charge from one point to another point so potential is a very very important Concept in electrostatics and uh as you can see here the electric field decreased as 1 by R squ the potential decreases as 1 by

00:48:24 R and there is a unit uh which is used for potential unit of potential is volt this is after an Italian scientist Alexandro vaa who was um who worked during the year 1745 to 1827 he was um he was a great scientist who did a lot of work in electrostatics and this name is this potential unit poten of potential is named after him so you can also then find another unit for electric field because as you can see here potential is Q by 40 r the electric field of the charge would have been Q by 40 R squ so you define

00:49:32 electric field as volt per meter this is the SI unit of um uh electric field volt per meter we have seen earlier other uh unit for electric field this is another standard unit which is used for electric field now let me calculate let me do an example to calculate some numerical values of potential so here is an example so I have um + 10 nanum I have another charge here minus 10 nanum separated by distance 6 cm in the equatorial plane I consider a point here uh this is 4 cm let me call this point

00:50:30 a I consider another Point here B this is at a distance of 4 cm from here and at third Point C which is U at a distance of 4 cm from here okay so I have + 10 NM minus 10 nanum and these two charges generate an electric field field I want to calculate the potential at different points so first let me start calculating potential calculation so V at a now potentials also satisfy principle of superposition so the potential at a total potential at a is the potential at a because of plus 10 nanum plus the

00:51:24 potential at a because ofus 10 nanum so if what is this distance if this is R because I'm taking a on the equatorial plane this distance is also R so potential at a is uh so let me call This q1 And Q2 q1 by 4 Pi Epsilon 0 R uh so plus Q2 by 4 pi0 R the distances are equal and Q2 is equal to minus q1 so this is equal to Z so the potential at this point is zero because that's an equatorial plane this produces a positive potential here that produces a negative potential here and so the total potential at this point

00:52:15 a is zero let me try to calculate the potential at the point B so V at B q1 by 4 Pi epon 0 so this I must calculate these distances now so let me call this uh R1 and the other distance R2 R1 minus Q2 is minus q1 so 4 pi0 R2 so let me substitute these numbers so V at B is uh 10 nanoc 1X 4 Pon 0 is 9 10^ 9 ID by R1 R1 is 4 cm so 4 into 10us 2 - 10 nanum 4 pepon 0 come here divided by R2 which is 4 + 6 10 cm please be careful with units I'm using SI units everywhere so this comes out to be 2.25 into 10^

00:53:37 3 minus 9 into 10^ 2 which is equal to 135 into 10^ 3 volts so that is the potential at this point so that is the work done in bringing a unit charge positive charge from Infinity to this point and that is equal to 1.35 10^3 volts you can similarly calculate the potential at the point C so V at C will be equal to so I leave this exercise to you this is - 1.25 into 10^ 3 Vol please note this point is closer to the negative charge here than the positive charge so the the overall potential happens to be zero minus 1.25

00:54:29 10 3 volts U and the potential at a was equal to Z now let me leave a problem to you problem calculate the work done in moving a charge of five nanocs from A to B and from a to c so I leave the exercise to you please calculate what is the work done in moving a charge of 5 nanum from A to B and from a to c I leave the paths to you to you choose appropriate paths to calculate the poent the the work done so it's an interesting exercise for you to understand calculation of uh calculation of forces and the

00:55:41 potential differences okay now what I would like to do is to look at the potential of a as another example potential of a charg conducting sphere so I have a sphere of radius R and I have thrown in some charge capital Q remember we have discussed that the entire charge will sit on the surface so I will end my lecture here in the next class what we will do is calculate the potential due to a conducting sphere which has been uh put excess charge and we'll get some interesting results from here thank you