Question

A solid sphere of radius R has a uniform charge density and total charge Q. Find the total energy of the sphere U in terms of

Answer 1

Answer:

$\mathbf{U = \dfrac{3 k_c Q^2_{total}}{5R}}$

Explanation:

From the information given;

The surface area of a sphere = $4 \pi r ^2$

If the sphere is from the collection of spherical shells of infinitesimal thickness = dr

Then,

the volume of the  thickness and the sphere is;

V = $4 \pi r ^2 \ dr$

Using Gauss Law

$V(r) = \dfrac{k_cq(r)}{r}$

here,

q(r) =charge built up contained in radius r

since we are talking about collections of spherical shells, to work required for the next spherical shell r +dr is

$-dW= dU = V(r) dq = \dfrac{k_c \ q(r)}{r} \ dq$

where;

$q (r) = \dfrac{4}{3} \pi r^3 \rho$

dq which is the charge contained in the  next shell of charge

here dq = volume of the shell multiply by the density

$dq = 4 \pi r^2 \ dr \ \rho$

equating it all together

$dU = \dfrac{k_c \frac{4}{3} \pi r^3 \rho}{r} 4 \pi r^2 \ dr \ \rho = \dfrac{16 \pi^2 \ k_c \ \rho^2}{3} \ r^4 \ dr$

Integration the work required from the initial radius r to the final radius R, we get;

$U = \int^R_0 \ dU$

$U = \int^R_0 \dfrac{16 \pi^2 \ k_c \rho^2}{3} r^4 \ dr$

$U = \int^R_0 \dfrac{16 \pi^2 \ k_c \rho^2}{3} [\dfrac{r^5}{5}]^R_0$

$U = \dfrac{16 \pi^2 k_c \rho^2}{15} \ R^5$

Recall that:

the total charge on a sphere, i.e $Q_{total} = \dfrac{4}{3} \pi R^3 \rho$

Then :

$\mathbf{U = \dfrac{3 k_c Q^2_{total}}{5R}}$