Question

A metal sphere of radius R is charged to a potential V. Find electrostatic energy stored in the electric field within a cocentric sphere of radius 2R. show that the electricstatic field energy stored outside the sphere of radius 2R equals energy stored inside it.

Answer 1

"Potential of the sphere,

V = q / 4πε0R

Here q is the charge on the sphere

1. The energy density at a distance r from the center of the sphere is,

Ue = ½ ε0E^2 = ½ ε0 (q/ 4πε0r^2) = q^2/ 32 π^2 ε0 r^4

If the spherical radius r and the thickness dr around the radius (r>R) then the spherical element is,

dV = 4 πr^2 dr

stored energy,

dUe = Ue * dV = (q^2/ 32 π^2 ε0 r^4) * 4 π r^2 dr = q^2/ 8 πε0 (dr/r^2)

on integrating it between R and 2R you will get,

q^2/ 16 πε0R

Since, q = 4 πε0RV

Ue = πε0RV^2

2. Integrate dUe between 2R and infinity

q = 4 πε0RV

therefore, U e = (4πε0RV)^2/ 16 πε0R

Ue = πε0RV^2

"

Answer 2

The metal sphere given in the question is given in the image below.

The electric potential is given by the formula:

$\bold{\mathrm{V}=\frac{q}{4 \pi \varepsilon_{\mathrm{o}} R}}$

Where,

q = Charge

R = Radius of the sphere

Determination of electrostatic energy:

The electrostatic energy is given by the formula:

$\bold{\mathrm{U}_{\mathrm{E}}=\frac{1}{2} \varepsilon_{0} \mathrm{E}^{2}}$

On substituting the electric field in the above equation, we get,

$\bold{U_E=\frac{1}{2} \varepsilon_{0}\left(\frac{q}{4 \pi \varepsilon_{0} r^{2}}\right)^{2}}$

$\bold{U_E=\frac{q^{2}}{32 \pi^{2} \varepsilon_{0} r^{4}}}$

Let's consider a element of radius r and thickness dr. The volume of the spherical element is:

$\bold{\mathrm{dV}=4 \pi \mathrm{r}^{2} \mathrm{dr}}$

The energy in the element is:

$\bold{\mathrm{dU}_{\mathrm{E}}=\mathrm{U}_{\mathrm{E}} \times \mathrm{d} \mathrm{V}}$

$\bold{dU_E=\frac{q^{2}}{32 \pi^{2} \varepsilon_{0} r^{4}} \times 4 \pi r^{2} d r}$

$\bold{dU_E=\frac{q^{2}}{8 \pi \varepsilon_{0}} \frac{d r}{r^{2}}\longrightarrow (1)}$

Now, the electrostatic energy in the sphere 2R is:

$\bold{U_{E}=\int_{R}^{2 R} \frac{q^{2}}{8 \pi \varepsilon_{0}} \frac{d r}{r^{2}}}$

$\bold{U_E=\frac{q^{2}}{8 \pi \varepsilon_{0}} \int_{R}^{2 R} \frac{d r}{r^{2}}}$

$\bold{U_E=\frac{q^{2}}{8 \pi \varepsilon_{0}}\left[-\frac{1}{r}\right]_{R}^{2 R}}$

$\bold{U_E=\frac{q^{2}}{16 \pi \varepsilon_{0} R}}$

Now, the charge is given by the formula:

$\bold{q=4 \pi \epsilon_0 R V}$

Now, the energy density becomes,

$\bold{U_{E}=\frac{q^{2}}{16 \pi \varepsilon_{0} R}}$

$\bold{U_E=\frac{\left(4 \pi \varepsilon_{0} R V\right)^{2}}{16 \pi \varepsilon_{0} R}}$

$\bold{\therefore\mathbf{U}_{\mathrm{E}}=\pi \varepsilon_{0} \mathrm{RV}^{2}}$

Electrostatic field energy:

Now, on integrating the equation (1), from 2R to infinity, we get,

$\bold{\mathrm{U}_{\mathrm{E}}=\int_{2 R}^{\infty} \frac{q^{2}}{8 \pi \varepsilon_{0}} \frac{d r}{r^{2}}}$

$\bold{U_E=\frac{q^{2}}{8 \pi \varepsilon_{0}} \int_{2 R}^{\infty} \frac{d r}{r^{2}}}$

$\bold{U_E=\frac{q^{2}}{8 \pi \varepsilon_{0}}\left[-\frac{1}{r}\right]_{2 R}^{\infty}}$

$\bold{U_E=\frac{q^{2}}{16 \pi \varepsilon_{0} R}}$

Now, on substituting the charge q, we get,

$\bold{q=4 \pi s_{0} R V}$

$\bold{U_{E}=\frac{q^{2}}{16 \pi \varepsilon_{0} R}}$

$\bold{U_E=\frac{\left(4 \pi \varepsilon_{0} R V\right)^{2}}{16 \pi \varepsilon_{0} R}}$

$\bold{\mathbf{U}_{\mathrm{E}}=\pi \varepsilon_{0} \mathrm{RV}^{2}}$

Thus, the electrostatic field energy stored outside the sphere of radius 2R equals energy stored inside it. Hence proved.