# 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

###### Question:

## Answers

[tex]\mathbf{U = \dfrac{3 k_c Q^2_{total}}{5R}}[/tex]

Explanation:

From the information given;

The surface area of a sphere = [tex]4 \pi r ^2[/tex]

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 = [tex]4 \pi r ^2 \ dr[/tex]

Using Gauss Law

[tex]V(r) = \dfrac{k_cq(r)}{r}[/tex]

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

[tex]-dW= dU = V(r) dq = \dfrac{k_c \ q(r)}{r} \ dq[/tex]

where;

[tex]q (r) = \dfrac{4}{3} \pi r^3 \rho[/tex]

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

here dq = volume of the shell multiply by the density

[tex]dq = 4 \pi r^2 \ dr \ \rho[/tex]

equating it all together

[tex]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[/tex]

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

[tex]U = \int^R_0 \ dU[/tex]

[tex]U = \int^R_0 \dfrac{16 \pi^2 \ k_c \rho^2}{3} r^4 \ dr[/tex]

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

[tex]U = \dfrac{16 \pi^2 k_c \rho^2}{15} \ R^5[/tex]

Recall that:

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

Then :

[tex]\mathbf{U = \dfrac{3 k_c Q^2_{total}}{5R}}[/tex]

Efficiency is an action of doing something. for example when you take a test. good efficiency means you did well but bad effeciency or no effiency means you did bad.

real, smaller and upside-down. so the answer should be b.