Question

An isolated conducting sphere of radius R=6.85cm has charge 1.25nC. Find the energy density at the surface of conducting
sphere.​

Answer 1

Answer:

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Answer 2

Answer:

The energy density at the surface of the conducting sphere is 1.1×10⁻⁷J/m².

Explanation:

The energy density at the surface of the sphere is given as,

$u=\frac{1}{2} \frac{Q^2}{A^2\epsilon_0}$       (1)

Where,

u=energy density of the sphere

Q=charge on the sphere

A=area of the sphere=4πr²     (r=radius of the sphere)

∈₀=permitivity of free space=8.854×10⁻¹²F.m⁻¹

From the question we have,

R=6.85cm=6.85×10⁻²m

Q=1.25nC=1.25×10⁻⁹C

By substituting all the required values in equation (1) we get;

$u=\frac{1}{2} \frac{(1.24\times 10^-^9)^2}{(4\pi\times (6.85\times 10^2)^2\epsilon_0}$

$u=1.1\times 10^-^7J/m^2$

Hence, the energy density at the surface of the conducting sphere is 1.1×10⁻⁷J/m².