TS EAMCET · Physics · Electrostatics
Two charges \(2 \mathrm{C}\) and \(6 \mathrm{C}\) are separated by a finite distance. If a charge of \(-4 \mathrm{C}\) is added to each of them, the initial force of \(12 \times 10^3 \mathrm{~N}\) will change to
- A \(4 \times 10^3 \mathrm{~N}\) (repulsion)
- B \(4 \times 10^2 \mathrm{~N}\) (repulsion)
- C \(6 \times 10^3 \mathrm{~N}\) (attraction)
- D \(4 \times 10^3 \mathrm{~N}\) (attraction)
Answer & Solution
Correct Answer
(D) \(4 \times 10^3 \mathrm{~N}\) (attraction)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Initial force, } F=k \frac{q_1 q_2}{r^2} \\ & F_1=k \cdot \frac{(2)(6)}{r^2} \\ & \text { New force, } F_2=k \frac{(2-4)(6-4)}{r^2}=k \frac{(-2)(2)}{r^2} \\ & \therefore \quad \frac{F_1}{F_2}=\frac{(2 \times 6)}{(-2 \times 2)} \\ & \frac{12 \times…
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