Question

A conducting sphere of radius $0.1\,\text{m}$ has uniform charge density $1.8\,\mu\text{C/m}^2$ on its surface. The electric field in free space at a radial distance $0.2\,\text{m}$ from the centre is ($\epsilon_0$ = permittivity of free space)

• A. $\frac{6\times10^{-6}}{\epsilon_{0}}Vm^{-1}$
• B. $\frac{6\times10^{-8}}{\epsilon_{0}}Vm^{-1}$
• C. $\frac{2\times10^{-7}}{\epsilon_{0}}Vm^{-1}$
• D. $\frac{1\times10^{-7}}{\epsilon_{0}}Vm^{-1}$

Hint:

Physics Tip : For any point outside a conducting sphere, the field behaves as if all the charge is concentrated at the geometric center. Always measure your "R" from the center, not the surface!

Answer 1

The Correct Option is C

Concept: Physics (Electrostatics) - Electric Field of a Spherical Conductor.

Step 1: Calculate the total charge on the sphere. The total charge $Q$ is the product of charge density $\sigma$ and surface area $A$: $$A = 4\pi r^{2} = 4\pi(0.1)^{2} = 0.04\pi~m^{2}$$ $$Q = \sigma \cdot A = (1.8\times10^{-6}) \cdot (0.04\pi) = 7.2\times10^{-8}\pi~C$$

Step 2: Determine the actual distance from the center. The distance is 0.2 m from the surface, so the total radial distance $R$ from the center is: $$R = 0.1~m + 0.2~m = 0.3~m$$

Step 3: Apply Gauss's Law for an external point. The electric field $E$ at distance $R$ is: $$E = \frac{Q}{4\pi\epsilon_{0}R^{2}} = \frac{7.2\times10^{-8}\pi}{\epsilon_{0} \cdot 4\pi(0.3)^{2}}$$ $$E = \frac{7.2\times10^{-8}}{4 \cdot \epsilon_{0} \cdot 0.09} = \frac{7.2\times10^{-8}}{0.36\epsilon_{0}}$$ $$E = 20\times\frac{10^{-8}}{\epsilon_{0}} = \frac{2\times10^{-7}}{\epsilon_{0}}Vm^{-1}$$ $$ \therefore \text{The electric field intensity is } \frac{2\times10^{-7}}{\epsilon_{0}}Vm^{-1}. $$