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ss7ja [257]
3 years ago
10

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

Physics
1 answer:
Lana71 [14]3 years ago
6 0

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}}

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