Question

Diameters of two spheres of metal are $6 \, \text{cm}$ and $4 \, \text{cm}$. They are charged to the same potential. Find out the ratio of the surface densities of charge on the sphere.

Hint:

For equal potentials, $\sigma \propto \dfrac{1}{R}$.

Answer 1

Step 1: Relation of potential with charge.
For a sphere of radius $R$, \[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R}. \] If both spheres are at same potential: \[ \frac{Q_1}{R_1} = \frac{Q_2}{R_2}. \]
Step 2: Express $Q$ in terms of surface density.
\[ Q = \sigma \cdot 4\pi R^2. \] So, \[ \frac{\sigma_1 \cdot 4\pi R_1^2}{R_1} = \frac{\sigma_2 \cdot 4\pi R_2^2}{R_2}. \] \[ \sigma_1 R_1 = \sigma_2 R_2. \]
Step 3: Substitution.
$R_1 = 3 \, \text{cm}$, $R_2 = 2 \, \text{cm}$. \[ \sigma_1 \cdot 3 = \sigma_2 \cdot 2. \] \[ \sigma_1 : \sigma_2 = 2 : 3. \]
Step 4: Conclusion.
The ratio of surface charge densities is $2 : 3$.