Question

Two conducting spheres of radii r1 and r2 are equally charged. The ratio of their potential is

• A. (r1/r2)²
• B. (r1r2)²
• C. (r1/r2)
• D. (r2/r1)

Answer 1

Answer: (D)

(r2/r1)

Answer 2

Let the radii of the two conducting spheres be $r_1$ and $r_2$. The problem states that the spheres are "equally charged". Let the charge on each sphere be $Q$.

The electric potential ($V$) on the surface of a conducting sphere of radius $r$ carrying a charge $Q$ is given by the formula:

$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r}$

For the first sphere with radius $r_1$ and charge $Q$, its potential $V_1$ is:

$V_1 = \frac{1}{4\pi\epsilon_0} \frac{Q}{r_1}$

For the second sphere with radius $r_2$ and charge $Q$, its potential $V_2$ is:

$V_2 = \frac{1}{4\pi\epsilon_0} \frac{Q}{r_2}$

To find the ratio of their potentials, we divide $V_1$ by $V_2$:

$\frac{V_1}{V_2} = \frac{\frac{1}{4\pi\epsilon_0} \frac{Q}{r_1}}{\frac{1}{4\pi\epsilon_0} \frac{Q}{r_2}}$

The terms $\frac{1}{4\pi\epsilon_0}$ and $Q$ are common in both the numerator and the denominator, so they cancel out:

$\frac{V_1}{V_2} = \frac{\frac{1}{r_1}}{\frac{1}{r_2}}$

$\frac{V_1}{V_2} = \frac{1}{r_1} \times r_2$

$\frac{V_1}{V_2} = \frac{r_2}{r_1}$

Thus, the ratio of their potentials is $(r_2/r_1)$.

Explanation: The potential of a conducting sphere is directly proportional to its charge and inversely proportional to its radius. Since both spheres have the same charge, their potential ratio is simply the inverse ratio of their radii.