Q159 / 200 · MHT CET - Medical 2026 (22 April Shift 1) · Physics
Two point charges $+3$ C and $+6$ C separated by a distance 'r' , repel each other with a force of 27 N. If each of the above charges is given an additional charge of $-5$ C, the force experienced by them is
- A$6\text{N}$ (attractive)
- B$6\text{N}$ (repulsive)
- Correct answerC$3\text{N}$ (attractive)
- D$3\text{N}$ (repulsive)
Solution
Initial charges are $q_1 = +3\text{ C}$ and $q_2 = +6\text{ C}$.
Initial force $F_1 = \dfrac{k q_1 q_2}{r^2} = \dfrac{k (3)(6)}{r^2} = \dfrac{18k}{r^2} = 27\text{ N}$.
This gives $\dfrac{k}{r^2} = \dfrac{27}{18} = 1.5\text{ N}$.
When an additional charge of $-5\text{ C}$ is given to each, the new charges become:
$q_1' = +3 - 5 = -2\text{ C}$
$q_2' = +6 - 5 = +1\text{ C}$
The new force $F_2 = \dfrac{k |q_1'| |q_2'|}{r^2} = \dfrac{k (2)(1)}{r^2} = \dfrac{2k}{r^2}$.
Substituting the value of $\dfrac{k}{r^2}$, we get:
$F_2 = 2 \times 1.5 = 3\text{ N}$.
Since the new charges have opposite signs, the force between them is attractive.
Answer: $3\text{N}$ (attractive)
Initial force $F_1 = \dfrac{k q_1 q_2}{r^2} = \dfrac{k (3)(6)}{r^2} = \dfrac{18k}{r^2} = 27\text{ N}$.
This gives $\dfrac{k}{r^2} = \dfrac{27}{18} = 1.5\text{ N}$.
When an additional charge of $-5\text{ C}$ is given to each, the new charges become:
$q_1' = +3 - 5 = -2\text{ C}$
$q_2' = +6 - 5 = +1\text{ C}$
The new force $F_2 = \dfrac{k |q_1'| |q_2'|}{r^2} = \dfrac{k (2)(1)}{r^2} = \dfrac{2k}{r^2}$.
Substituting the value of $\dfrac{k}{r^2}$, we get:
$F_2 = 2 \times 1.5 = 3\text{ N}$.
Since the new charges have opposite signs, the force between them is attractive.
Answer: $3\text{N}$ (attractive)
Hence, the correct answer is option: 'C'.