COMEDK · Physics · 17. Electrostatics
Two point charges \(A=+3 \mathrm{nC}\) and \(B=+1 \mathrm{nC}\) are placed \(5 \mathrm{~cm}\) apart in air. The work done to move charge \(B\) towards \(A\) by \(1 \mathrm{~cm}\) is
- A \(20 \times 10^{-7} \mathrm{~J}\)
- B \(1.35 \times 10^{-7} \mathrm{~J}\)
- C \(27 \times 10^{-7} \mathrm{~J}\)
- D \(121 \times 10^{-7} \mathrm{~J}\)
Answer & Solution
Correct Answer
(B) \(1.35 \times 10^{-7} \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
The given situation is shown below \[ \begin{aligned} r_{1} &=1 \mathrm{~cm} \\ &=1 \times 10^{-2} \mathrm{~m} \\ r_{2} &=5-1 \\ &=4 \mathrm{~cm} \\ &=4 \times 10^{-2} \mathrm{~m} \end{aligned} \] Required work done \(=\) Changes in potential energy…
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