Question

Coulomb's repulsive force of \( F \) newton acts between two point charges of \( +3\,\mu C \) and \( +8\,\mu C \). If \( -5\,\mu C \) charge is given to both of them, then the force and its nature between them will be:

• A. \( F/4 \) newton, attractive
• B. \( F/4 \) newton, repulsive
• C. \( 3F/13 \) newton, repulsive
• D. \( 4F \) newton, attractive

Hint:

Update each charge by adding \( -5\,\mu C \); force scales as the product of charges, and opposite signs mean attraction.

Answer 1

The Correct Option is A

Step 1: Since the distance is unchanged, the force is proportional only to the product of the two charges: \(F \propto q_1 q_2\).

Step 2: Original product \(= (+3)(+8) = +24\) (in \(\mu C\)), a positive product meaning a repulsive force of value \(F\).

Step 3: After sharing \(-5\,\mu C\) to each plate, the charges become \(-2\,\mu C\) and \(+3\,\mu C\). Their product \(= (-2)(+3) = -6\). The magnitude is 6 and the negative sign tells us the force is attractive.

Step 4: Ratio of magnitudes \(= \dfrac{6}{24} = \dfrac{1}{4}\), so the new force is one quarter of \(F\) and attractive.

\[\boxed{F' = \frac{F}{4}\ \text{, attractive}}\]