Question

Two spherical conductors of radii $4\text{ cm}$ and $5\text{ cm}$ are charged to the same potential. If '$\sigma_1$' and '$\sigma_2$' be the respective values of the surface density of charge on the two conductors then the ratio $\sigma_1 : \sigma_2$ is

• A. $5 : 4$
• B. $3 : 2$
• C. $4 : 3$
• D. $2 : 1$

Hint:

For spheres at equal potential, the surface charge density is inversely proportional to the radius ($\sigma \propto \frac{1}{R}$). Therefore, you can find the density ratio by simply flipping the radius values: $\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1} = \frac{5}{4}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given two isolated spherical conductors with different radii, $R_1 = 4\text{ cm}$ and $R_2 = 5\text{ cm}$. Both spheres are charged until they reach the exact same electric potential ($V_1 = V_2$). We need to determine the ratio of their surface charge densities, $\sigma_1 : \sigma_2$.

Step 2: Key Formula or Approach:
The electric potential $V$ on the surface of a charged conducting sphere of radius $R$ containing total charge $Q$ is: $$V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q}{R}$$ The surface charge density $\sigma$ is defined as the total charge divided by the surface area of the sphere: $$\sigma = \frac{Q}{4\pi R^2} \implies Q = \sigma(4\pi R^2)$$ Substituting this charge definition into the potential equation reveals how potential relates to surface density: $$V = \frac{1}{4\pi\varepsilon_0} \cdot \frac{\sigma(4\pi R^2)}{R} = \frac{\sigma R}{\varepsilon_0}$$

Step 3: Detailed Explanation:
Since both spheres are at the same electric potential, we can equate their potential expressions: $$V_1 = V_2 \implies \frac{\sigma_1 R_1}{\varepsilon_0} = \frac{\sigma_2 R_2}{\varepsilon_0}$$ The constant term $\varepsilon_0$ cancels out from both sides of the equation: $$\sigma_1 R_1 = \sigma_2 R_2$$ Rearrange this equation to find the ratio of the surface densities: $$\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}$$ Substitute the given values for the radii ($R_1 = 4\text{ cm}$ and $R_2 = 5\text{ cm}$): $$\frac{\sigma_1}{\sigma_2} = \frac{5}{4}$$ This gives a final ratio of $5 : 4$.

Step 4: Final Answer:
The ratio of the surface charge densities is $5 : 4$, which corresponds to option (A).