Question

A spherical balloon of radius r charged uniformly on it

Answer 1

Look at the potentials due to the sphere of radius R and radius 2R:

Start with electric field sing Gauss' law:

E(R) = 4*pi*R^2*sigm/(e0*4*pi*r^2) = sigm*R^2/(e0*r^2) for r > R

Now assuming you don't chreate more charge as the balloon expands then the electric field for the radius 2R is the same as for R:

Q = sigm*4*pi*R^2 ---> E(2R) = Q/(4*pi*e0*r^2) = sigm*R^2/(e0*r^2) r > 2R

The potentials for each field are given by:

V = -integral(E dr) --> V(R) = -integral[R, infinity](sigm*R^2/(e0*r^2) dr) = -sigm*R^2/(e0*R) = -sigm*R/e0

By analogy then V(2R) = -sigm*(R)^2/(e0*2R) = -sigm*R/(2e0) = V(R)/2

If the work = charge time change in potential = change in potential energy then

W = Q*(V(R) - V(2R)) = Q*V(R)/2 = sigm*4*pi*R^2*(-sigm*R/(e0*2)) = -2*pi*sigm^2*R^3/e0

Answer 2

Answer:

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