Question

Two point charges +6 \(\mu\)C and +10 \(\mu\)C kept at a certain distance repel each other with a force of 30 N. If each charge is given an additional charge of -8 \(\mu\)C, the two charges

• A. Attract with a force of 2N
• B. Repel with a force of 2N
• C. Attract with a force of 15N
• D. Repel with a force of 15N

Hint:

If both charges remain positive or negative, they repel. If one becomes negative, they attract.

Answer 1

The Correct Option is A

Let the initial charges be \( q_1 = +6 \mu C \) and \( q_2 = +10 \mu C \).
The initial force of repulsion is \( F_1 = 30 N \).
After adding \(-8 \mu C\) to each charge, the new charges are:
\[ q_1' = +6 \mu C - 8 \mu C = -2 \mu C \] \[ q_2' = +10 \mu C - 8 \mu C = +2 \mu C \]
The initial force is given by Coulomb's law: \[ F_1 = k \frac{q_1 q_2}{r^2} \] where \( k \) is Coulomb's constant and \( r \) is the distance between the charges.
The new force is given by: \[ F_2 = k \frac{q_1' q_2'}{r^2} \]
We can write: \[ \frac{F_2}{F_1} = \frac{k \frac{q_1' q_2'}{r^2}}{k \frac{q_1 q_2}{r^2}} = \frac{q_1' q_2'}{q_1 q_2} \]
Substituting the values: \[ \frac{F_2}{30 N} = \frac{(-2 \mu C)(+2 \mu C)}{(+6 \mu C)(+10 \mu C)} = \frac{-4}{60} = -\frac{1}{15} \]
So, \[ F_2 = 30 N \times \left( -\frac{1}{15} \right) = -2 N \]
The negative sign indicates that the force is attractive.
The magnitude of the force is \( |F_2| = 2 N \).
Therefore, the two charges attract each other with a force of 2 N.