Question

Force between two charges \( +8\,\mu C \) and \( +2\,\mu C \) is \( 16\,N \). If the charges are brought into contact and then separated by the same distance, the force between them is

• A. \( 25\,N \)
• B. \( 20\,N \)
• C. \( 30\,N \)
• D. \( 15\,N \)
• E. \( 12\,N \)

Hint:

When identical conductors touch, charges redistribute equally. Always find new charges first, then apply Coulomb’s law again.

Answer 1

The Correct Option is A

Step 1: Recall Coulomb’s law.
The force between two charges is given by: \[ F = k \frac{q_1 q_2}{r^2} \] where \( q_1, q_2 \) are charges and \( r \) is the distance between them.

Step 2: Write the initial condition.
Initially: \[ q_1 = 8\,\mu C,\quad q_2 = 2\,\mu C \] Given force: \[ F_1 = 16\,N \] So, \[ F_1 = k \frac{(8)(2)}{r^2} = k \frac{16}{r^2} \]

Step 3: Find the new charges after contact.
When two conductors are brought into contact, charge redistributes equally.
Total charge: \[ q_{\text{total}} = 8 + 2 = 10\,\mu C \] After sharing equally: \[ q_1' = q_2' = \frac{10}{2} = 5\,\mu C \]

Step 4: Write the new force expression.
After separation at same distance: \[ F_2 = k \frac{(5)(5)}{r^2} = k \frac{25}{r^2} \]

Step 5: Take ratio of new and initial force.
\[ \frac{F_2}{F_1} = \frac{25}{16} \]

Step 6: Calculate the new force.
\[ F_2 = 16 \times \frac{25}{16} = 25\,N \]

Step 7: Final conclusion.
Hence, the new force is: \[ \boxed{25\,N} \] Therefore, the correct option is \[ \boxed{(1)\ 25\,N} \]