Question

There are two charged spheres of radius $R$ and $3R$. When the spheres are made to touch each other and then separate, the surface charge density becomes $r_1$ and $r_2$ respectively. Find $\frac{r_1}{r_2}$.

• A. \( \frac{1}{9} \)
• B. \( \frac{1}{3} \)
• C. 3
• D. 9

Hint:

When two charged spheres of different sizes touch, the surface charge density on each is inversely proportional to the square of the radius.

Answer 1

The Correct Option is A

When two charged spheres touch each other, charge redistributes between them. The surface charge density \( \sigma \) is given by: \[ \sigma = \frac{Q}{A}, \] where \( Q \) is the charge on the sphere and \( A \) is the surface area of the sphere.
- The charge on a sphere is proportional to its radius, i.e., \( Q \propto R^2 \), where \( R \) is the radius of the sphere.
- When the spheres are in contact, they share charge. The total charge on the system is conserved, but the redistribution of charge depends on the radii of the spheres.
Let the charges on the spheres be \( Q_1 \) and \( Q_2 \), corresponding to radii \( R \) and \( 3R \), respectively. Since charge is proportional to the surface area, we can express the charge densities after they touch as: \[ r_1 = \frac{Q_1}{4 \pi R^2}, \quad r_2 = \frac{Q_2}{4 \pi (3R)^2}. \] The total charge is conserved, and since the radii of the spheres are different, the surface charge density will be inversely proportional to the square of the radius. Using this relationship: \[ \frac{r_1}{r_2} = \frac{Q_1 / (4 \pi R^2)}{Q_2 / (4 \pi (3R)^2)} = \frac{Q_1 / R^2}{Q_2 / 9R^2} = \frac{Q_1}{Q_2} \times 9. \] Since the total charge is conserved, the ratio \( \frac{Q_1}{Q_2} = \frac{1}{9} \), giving: \[ \frac{r_1}{r_2} = \frac{1}{9}. \] Thus, the correct answer is (1) \( \frac{1}{9} \).