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Can anyone explain the derivation for the expression of the electric potential INSIDE a non conducting solid sphere?

Can anyone explain the derivation for the expression of the electric potential INSIDE a non conducting solid sphere?

Grade:12th Pass

1 Answers

Arun
25750 Points
4 years ago

The electric potential is defined as negative of the work done by the field on unit positive charge in order to bring it from infinity( reference position) to the point where the potential is to be found.

As far as the point is outside the sphere or on the surface of the sphere, the potential is such that as if the whole charge in the sphere is concentrated at the center of the sphere. Therefore, the potential at any point on the surface of the sphere is kQ/R, where Q is total charge in the sphere of radius R. k is constant appearing in Coulomb's law.

Now, suppose ,we want to find potential at distance r1(

Now, total charge in the sphere is Q. The charge density will be d=Q/[(4/3)pi R^3]. Therefore , charge in the sphere of radius r willbe

[(4/3)pi r^3]d=[4/3) pi r^3]Q/[(4/3) pi R^3=Q r^3/R^3. Applying Gauss's theorem, the field at r is

E(r)= kQr/R^3. Therefore,

negative of integral from R to r1 is -kQ/R^3[r^2/2] from R to r1=-kQ/R^3[r1^2/2-R^2/2]=-(1/2)(kQ)(r1^2/R^3) +(kQ)(1/2)(1/R)

=-(1/2)(kQ)[r1^2/R^3-(1/2)R].

This result added to kQ/R gives potential at r1. Therefore ,

V(r1)=(1/2)kQ/R-(1/2)kQr1^2/R^3.=[(1/2)kQ/R] [1-(r1/R)^2].

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