Question

Two charged metallic spheres are joined by a very thin metal wire. If the radius of the larger sphere is four times that of the smaller sphere, the electric field near the larger sphere is

• A. twice that near the smaller sphere
• B. quarter of that near the smaller sphere
• C. same as that near the smaller sphere
• D. half of that near smaller sphere

Hint:

For connected conductors at potential equilibrium, the surface electric field is inversely proportional to the radius of curvature ($E \propto \frac{1}{R}$). Since the larger sphere has 4 times the radius, its surface electric field must drop to exactly $\frac{1}{4}$ of the smaller sphere's field.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
Two conducting spheres of different radii are linked by a thin conducting wire. This wire enables charge distribution until equilibrium is reached. We need to compare the resulting electric field intensities at their surfaces given that $R_{\text{large}} = 4R_{\text{small}}$.

Step 2: Key Formula or Approach:
Connecting two conductors with a wire forces them to share charges until their electrical potential fields balance perfectly ($V_{\text{large}} = V_{\text{small}}$). 1. The electric potential on a sphere's surface is:
$$V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$$ 2. The electric field strength near a sphere's surface is:
$$E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}$$ Combining these equations yields a direct link between surface field intensity and electric potential:
$$E = \frac{V}{R}$$

Step 3: Detailed Explanation:
Let the properties of the smaller sphere be $R_s$ and $E_s$, and the properties of the larger sphere be $R_l$ and $E_l$. We are given that $R_l = 4R_s$. Because they are joined by a conducting wire, their final potentials are equal:
$$V_s = V_l = V$$ Write down the electric field expressions using this shared potential equilibrium state:
$$E_s = \frac{V}{R_s}$$ $$E_l = \frac{V}{R_l}$$ Take the ratio of the large sphere's field to the small sphere's field:
$$\frac{E_l}{E_s} = \frac{\frac{V}{R_l}}{\frac{V}{R_s}} = \frac{R_s}{R_l}$$ Substitute $R_l = 4R_s$ into this ratio equation:
$$\frac{E_l}{E_s} = \frac{R_s}{4R_s} = \frac{1}{4}$$ $$E_l = \frac{1}{4}E_s$$ This means the surface electric field intensity of the larger sphere is exactly one-quarter of the field strength found near the smaller sphere.

Step 4: Final Answer:
The electric field near the larger sphere is a quarter of that near the smaller sphere, matching option (B).