Question

Two spherical conductors B and C having equal radii and carrying equal charges in them repel each other with a force $F$, when kept apart at some distance. A third spherical conductor having same radius as that of B but uncharged, is brought in contact with B, then brought in contact with C and finally removed away from both. The new force of repulsion between B and C is:

• A. $\frac{F}{4}$
• B. $\frac{3F}{4}$
• C. $\frac{F}{8}$
• D. $\frac{3F}{8}$

Answer 1

The correct option is D $\frac{3F}{8}$

$\text{Step 1: Calculation of initial force}$

Let $B$ and $C$ have charge $Q$ each initially and are seperated by $r$ distance.

From Coulomb's law, force between them $F=\frac{K{Q}^2}{r^2}$ $....\left(1\right)$

$\text{Step 2: Distribution of charges}$

When an identical uncharged spherical conductor $A$ is brought in contact with charged sphere $B$.

By symmetry, the total charge will be shared equally among them, as both have equal radii.

$\therefore$ Charge on $A$ and $B$ $=\frac{Q}{2}$

Now, when the conductor $A$ is brought in contact with $C$.

They will also share equal charge among themselves, as both have equal radii.

$\therefore$ Charge on $A$ and $C$ $=\frac{Q+\frac{Q}{2}}{2}=\frac{3Q}{4}$

$\text{Step 3: Calculation of new force}$

New coulomb force between $B$ and $C$:

$F^{\prime }=\frac{K\frac{Q}{2}\times \frac{3Q}{4}}{r^2}=\frac{3}{8}\frac{K{Q}^2}{r^2}$

From equation $\left(1\right)$, we get: $F^{\prime }=\frac{3}{8}F$

Hence, Option D is correct