Question

Two metal spheres, one of radius $ \frac{R}{2} $ and the other of radius $ 2R $ respectively have the same surface charge density. They are brought in contact and separated. The ratio of their new surface charge densities is:

• A. 2 : 1
• B. 4 : 1
• C. 1 : 4
• D. 1 : 2

Hint:

When two conducting spheres with different radii are brought into contact, charge redistributes between the spheres in proportion to their radii squared. This principle can be applied when solving charge distribution problems.

Answer 1

The Correct Option is B

The surface charge density \( \sigma \) of a sphere is defined as: \[ \sigma = \frac{Q}{4 \pi r^2} \] Where \( Q \) is the charge and \( r \) is the radius. Initially, the two spheres have the same surface charge density, so we have: \[ \sigma_1 = \sigma_2 \] Let \( Q_1 \) and \( Q_2 \) be the charges on the spheres initially. Since the surface charge densities are the same: \[ \frac{Q_1}{4 \pi \left( \frac{R}{2} \right)^2} = \frac{Q_2}{4 \pi \left( 2R \right)^2} \] Simplifying this: \[ \frac{Q_1}{\left( \frac{R}{2} \right)^2} = \frac{Q_2}{(2R)^2} \] \[ Q_1 = \frac{Q_2 \cdot R^2}{4R^2} \] So, \( Q_1 = \frac{Q_2}{4} \). When the spheres are brought into contact, the total charge \( Q_1 + Q_2 \) is shared between the two spheres. The charge will distribute according to their radii: \[ \frac{Q_1'}{Q_2'} = \frac{r_1}{r_2} = \frac{R/2}{2R} = \frac{1}{4} \] 
Thus, the new surface charge densities are: \[ \sigma_1' = \frac{Q_1'}{4 \pi \left( \frac{R}{2} \right)^2}, \quad \sigma_2' = \frac{Q_2'}{4 \pi (2R)^2} \] Therefore, the ratio of the new surface charge densities is: \[ \frac{\sigma_1'}{\sigma_2'} = \frac{4}{1} \] Thus, the new surface charge density ratio is 4 : 1.