Electrostatic Force on Conducting Spheres PYQ | JEE Main Physics 2025

 

❓ Question

Two identical point charges

$$+q \qquad\text{and}\qquad +q$$

are separated by a distance

$$d.$$

The point charges are now replaced by identical conducting spheres, each carrying charge $+q$, while the distance between their centers remains $d$.

Compare the electrostatic repulsive force before and after replacement.

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✍️ Solution

Step 1: Initial force (Point Charges)

For two point charges,

$$F=\frac{kq^2}{d^2}.$$

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Step 2: After replacing by conducting spheres

When the charges are placed on conducting spheres, the charges redistribute due to electrostatic induction.

• More positive charge accumulates on the outer surfaces.
• Less positive charge remains on the facing surfaces.

Hence, the electric field is not the same as that of point charges.

The effective force acts as if the charges are separated by a distance greater than $d$.

If the effective shift of charge centers is $r_1$ and $r_2$, then the effective separation becomes

$$d+r_1+r_2.$$

Therefore,

$$F' \propto \frac{1}{(d+r_1+r_2)^2}.$$

Since

$$d+r_1+r_2>d,$$

we have

$$\frac{1}{(d+r_1+r_2)^2} < \frac{1}{d^2}.$$

Thus,

$$F' < F.$$

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✅ Final Answer

$$\boxed{F' < F}$$

The repulsive force decreases when the point charges are replaced by identical conducting spheres carrying the same charge.