MHT CET · Physics · Electrostatics
Two point charges \(q_1\) and \(q_2\) are ' \(l\) ' distance apart. If one of the charges is doubled and the distance between them is halved, the magnitude of force becomes \(n\) times, where \(n\) is
- A \(8\)
- B \(1\)
- C \(2\)
- D \(16\)
Answer & Solution
Correct Answer
(A) \(8\)
Step-by-step Solution
Detailed explanation
The correct option is (A)
Concept: Functional dependence of Coulombic Force on charges \(\left(q_1, q_2\right)\) separated with distance \(l\) is given by
\(F \propto \frac{q_1 q_2}{l^2}\)
Therefore, if one of the charges is doubled and the distance between them is halved, the magnitude of new force
\(F_{\text {new }} \propto\left\{\frac{2 q_1 q_2}{\left(\frac{l}{2}\right)^2}\right\}\)
So, \(F_{\text {new }}=8 F\).
Concept: Functional dependence of Coulombic Force on charges \(\left(q_1, q_2\right)\) separated with distance \(l\) is given by
\(F \propto \frac{q_1 q_2}{l^2}\)
Therefore, if one of the charges is doubled and the distance between them is halved, the magnitude of new force
\(F_{\text {new }} \propto\left\{\frac{2 q_1 q_2}{\left(\frac{l}{2}\right)^2}\right\}\)
So, \(F_{\text {new }}=8 F\).
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