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: Understand the connection.
Two spheres joined by a thin wire share charge until they reach the same potential. The larger sphere has radius $R_l = 4R_s$. We compare the surface electric fields.
Step 2: Write the potential of a sphere.
$V = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}$, and the wire forces $V_l = V_s$.
Step 3: Write the surface field of a sphere.
$E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R^2}$.
Step 4: Relate field to potential.
Dividing the field by the potential gives $E = \dfrac{V}{R}$, a neat link valid for each sphere.
Step 5: Take the ratio with equal potentials.
$\dfrac{E_l}{E_s} = \dfrac{V/R_l}{V/R_s} = \dfrac{R_s}{R_l}$.
Step 6: Insert the radius relation.
$\dfrac{E_l}{E_s} = \dfrac{R_s}{4R_s} = \dfrac{1}{4}$, so the larger sphere's field is a quarter of the smaller sphere's. This is option (2).
\[ \boxed{E_l = \tfrac{1}{4}E_s} \]