A solid conducting sphere has net positive charge and radius...

A solid conducting sphere has net positive charge and radius R = 0.500 m . At a point 1.20 m from the center of the sphere, the electric potential due to the charge on the sphere is 27.0 V . Assume that V = 0 at an infinite distance from the sphere. a)What is the electric potential at the center of the sphere? Express your answer with the appropriate units.

Transcript

00:01 Hello leonard, it solved this problem.

00:02 Please consider the schematic diagram on the screen a positively charged solar sphere with radius 0 .50 meter and it goes a potential of 27 volt at a point a which is at a distance 1 .2 meter away from the center of the solid sphere.

00:29 Well before we saw this problem we have to know what is the base what is the potential what's what is the concept called electric potential at a point we have to know this then only we can solve this problem okay the potential at any point the potential at any point is the work done is the work done in bringing unit positive charge from infinity to that point the work done in bringing a unit positive charge from infinity to that point is the potential at that point.

01:09 So the potential at a we have given that with bas who go to 27 mode.

01:16 To 27mote is the work done in bringing a unit positive charge from infinity this point because this work done against the electric field impact here because this solid sphere will be causing electric field around okay so to bring a unity positive charge in an electric field of any charge system we have to do a work that work is regarded as the potential so more and more work work should be done to bring charge near to the sphere because the field will be more so nearing the surface the work done to bring a positive charge door will be more and more which means near the surface surface or nearly when we approach, when we go near to the charger system, the potential should be more on the surface, the potential be maximum.

02:24 Okay.

02:25 But then the twist happens.

02:29 As i said, the work done is against the field impact of electric field or the field impact of the charge to system but inside when we go inside as we said more more and work done and keep on doing and we reached to the surface but from the what happened is like that there needn't be more work and additional work then to bring the chart inside the sphere because inside the sphere the field is zero so what will be the work done to reach here the surface that work is enough to bring it into the center or inside a holosphere or solid sphere this concept we have to keep in mind okay which means what will be the potential at the surface the same potential will be there every world inside the sphere so back to the question what is the potential at the center the potential at the center is equal to the potential on the surface.

03:43 Vo, the potential at the origin is equal to the potential on the surface vr.

03:50 Okay, so let's consider vr, then we can find it very easy.

03:56 Vr, the potential on the surface, is equal to 1x4 by epsilon 0, q by, okay.

04:13 A v a is equal to 1 by 4 pi epsilon 0 q divided by distance is 1 .2 1 .2 is equal to 27 volt we already given with so that 1 by 4 pi epsilon 0 q is equal to we are going to take that 1 .2 to the other side 1 .4 by epsilon q is equal to 27 into 1 .2.

04:55 207 .1 .2...

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