© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Problem
We have two identical spherical conductors, say $B$ and $C$, each carrying the same charge $Q$. They are placed at a certain distance apart and repel each other with force $F$. A third identical sphere (initially uncharged) is successively brought into contact with $B$, then with $C$, and finally removed. We want to find the new force of repulsion between $B$ and $C$ after this process.
Step 2: Recall Coulomb’s Law
The initial repulsive force $F$ between two uniformly charged spheres carrying charges $Q_B$ and $Q_C$ separated by a distance $x$ is given by:
$$F \;=\; k\,\frac{Q_B\,Q_C}{x^2}$$
where $k$ is a constant (Coulomb constant). As long as distance $x$ is unchanged, the force is proportional to the product of charges, i.e.,
$$ F \propto \frac{Q_B\,Q_C}{x^2}.$$
Step 3: Charge Distribution When the Third Sphere Touches $B$
The third sphere is uncharged initially and has the same radius as $B$. When it is brought into contact with $B$, the charges redistribute equally because the spheres are identical. Initially, $B$ has charge $Q$ and the third sphere has charge $0$. After contact:
Charge on $B$ becomes $Q/2$.
Charge on the third sphere also becomes $Q/2$.
Step 4: Charge Distribution When the Third Sphere Touches $C$
Now the same third sphere (carrying $Q/2$) is brought into contact with $C$, which has charge $Q$. Both spheres again share the total charge equally (they are identical spheres). The total charge before contact is:
$$ Q + \frac{Q}{2} = \frac{3Q}{2}. $$
Since they share equally, each sphere gets:
$$ \text{Charge on each} = \frac{\frac{3Q}{2}}{2} = \frac{3Q}{4}. $$
Thus, after contact with $C$:
Charge on $C$ becomes $\tfrac{3Q}{4}$.
Charge on the third sphere becomes $\tfrac{3Q}{4}$.
Step 5: Final Charges on $B$ and $C$ After the Third Sphere is Removed
The important detail is that $B$ is no longer touched after it received charge $Q/2$. It remains with $Q/2$. Meanwhile, $C$ remains with $3Q/4$ after the contact with the third sphere. Now the third sphere is removed from the system.
Step 6: Calculate the New Force of Repulsion
With $B$ having charge $Q/2$ and $C$ having charge $3Q/4$, and the distance $x$ unchanged, the new force $F_{\text{new}}$ is:
$$F_{\text{new}} \;=\; k\,\frac{\left(\frac{Q}{2}\right)\left(\frac{3Q}{4}\right)}{x^2}
\;=\; k\,\frac{\frac{3Q^2}{8}}{x^2}
\;=\;\frac{3}{8}\,k\,\frac{Q^2}{x^2}.$$
But we know from the initial situation that
$$ F \;=\; k\,\frac{Q^2}{x^2}. $$
Therefore,
$$F_{\text{new}} = \frac{3}{8} F.$$
Step 7: State the Final Answer
The new force of repulsion between $B$ and $C$ is:
$$ \boxed{\frac{3F}{8}.} $$
