Question

Two equally charged small balls placed at a fixed distance experience a force 'F'. A similar uncharged ball after touching one of them is placed at the middle point between the two balls. The force experienced by this ball is ______.

• A. $F/2$
• B. $F$
• C. $2F$
• D. $4F$

Hint:

When identical conducting spheres touch, they behave as a single system. The total charge is summed up and then divided equally by 2 upon separation.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This problem applies Coulomb's Law and the principle of charge sharing by conduction. We must track the charges as they change and then calculate the net electrostatic force on the third ball placed in the middle.

Step 2: Detailed Explanation:
Let the initial two identical balls be A and B, separated by a distance $r$.
They are equally charged, so $q_A = q$ and $q_B = q$.
The initial force between them is $F = \frac{1}{4\pi\epsilon_0} \frac{q^2}{r^2} = k\frac{q^2}{r^2}$.
Now, an uncharged identical ball C ($q_C = 0$) touches ball A.
Because they are identical, they share the total charge equally.
New charge on A: $q'_A = \frac{q + 0}{2} = \frac{q}{2}$.
Charge on C: $q'_C = \frac{q}{2}$.
Ball C is now placed exactly at the midpoint between A and B.
Distance from C to A = $r/2$. Distance from C to B = $r/2$.
We must calculate the net force on C.
Force on C due to A ($F_{CA}$): Both are positive, so A repels C to the right.
$$F_{CA} = k \frac{q'_A q'_C}{(r/2)^2} = k \frac{(q/2)(q/2)}{r^2/4} = k \frac{q^2/4}{r^2/4} = k\frac{q^2}{r^2} = F$$
Force on C due to B ($F_{CB}$): Both are positive, so B repels C to the left.
$$F_{CB} = k \frac{q_B q'_C}{(r/2)^2} = k \frac{(q)(q/2)}{r^2/4} = k \frac{q^2/2}{r^2/4} = 2k\frac{q^2}{r^2} = 2F$$
Because these two repulsive forces point in opposite directions, we subtract them to find the net force on C.
$$F_{net} = F_{CB} - F_{CA} = 2F - F = F$$
The net force points toward A.

Step 3: Final Answer:
The force experienced by the middle ball is $F$, matching option (b).