AP EAMCET · PHYSICS · Electrostatics
A charge of \(1 \mu \mathrm{C}\) is divided into two parts such that their charges are in the ratio of \(2: 3\). These two charges are kept at a distance \(1 \mathrm{~m}\) apart in vacuum. Then, the electric force between them \((\) in \(N)\) is
- A \(0.216\)
- B \(0.00216\)
- C \(0.0216\)
- D \(2.16\)
Answer & Solution
Correct Answer
(B) \(0.00216\)
Step-by-step Solution
Detailed explanation
Ratio of charges \(=2: 3\) \[ \therefore q_1=\frac{2}{5} \times 1 \mu \mathrm{C} \text { and } q_2=\frac{3}{5} \times 1 \mu \mathrm{C} \] Electrostatic force between the two charges…
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