Question

Two charged identical metal spheres A and B repel each other with a force \(3\times 10^{-5}\) N. Another identical uncharged sphere C is touched with A and then placed at the mid-point between A and B. Net force on C is \(x\times 10^{-5}\) N. The value of \(x\) is

• A. \(1\)
• B. \(2\)
• C. \(1.5\)
• D. \(3\)

Hint:

Charge sharing halves the charge on A and gives C the same. Then find the forces from A and from B on C at the midpoint and subtract.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Identical spheres share charge equally on contact. Initially \(F=\frac{kq^{2}}{d^{2}}=3\times10^{-5}\ \text{N}\).

Step 2: Charge After Touching
A and C each carry \(\frac{q}{2}\). B still has \(q\).

Step 3: Forces on C
At distance \(\frac{d}{2}\):
From A: \(\frac{k(q/2)(q/2)}{(d/2)^{2}}=\frac{kq^{2}}{d^{2}}=F\) (away from A).
From B: \(\frac{k(q/2)(q)}{(d/2)^{2}}=\frac{2kq^{2}}{d^{2}}=2F\) (away from B).

Step 4: Net Force
The forces are opposite, so net \(=2F-F=F=3\times10^{-5}\ \text{N}\), giving \(x=3\).

Final Answer:
The net force on C is 3 times 10 to the minus 5 newton, so x is 3. \[ \boxed{\text{(D) }3} \]