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: Coulomb's law gives the force between two point charges as \[ F = \frac{1}{4\pi\varepsilon_0}\,\frac{q_1 q_2}{r^2}. \] The separation \( r \) stays the same throughout.

Step 2: Initial charges are \( q_1 = +3\,\mu C \) and \( q_2 = +8\,\mu C \), so \[ F = \frac{1}{4\pi\varepsilon_0}\,\frac{(3)(8)}{r^2} = \frac{1}{4\pi\varepsilon_0}\,\frac{24}{r^2}\ \text{(in }\mu C \text{ units).} \]

Step 3: Adding \( -5\,\mu C \) to each charge: \[ q_1' = 3 + (-5) = -2\,\mu C, \qquad q_2' = 8 + (-5) = +3\,\mu C. \]

Step 4: New force magnitude \[ F' = \frac{1}{4\pi\varepsilon_0}\,\frac{|(-2)(3)|}{r^2} = \frac{1}{4\pi\varepsilon_0}\,\frac{6}{r^2}. \]

Step 5: Take the ratio \[ \frac{F'}{F} = \frac{6}{24} = \frac{1}{4} \ \Rightarrow\ F' = \frac{F}{4}. \] The new charges have opposite signs \( (-2\,\mu C \) and \( +3\,\mu C) \), so the force is attractive. Hence option 1.

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