AP EAMCET · PHYSICS · Electrostatics
Two equally charged metal spheres \(A\) and \(B\) repel each other with a force of \(4 \times 10^{-5} \mathrm{~N}\). Another identical uncharged sphere \(C\) is touched to \(A\) and then placed at the mid- point of the line joining the spheres \(A\) and \(B\). The net electric force on the sphere \(C\) is
- A \(4 \times 10^{-5} \mathrm{~N}\) from \(C\) to \(A\)
- B \(4 \times 10^{-5} \mathrm{~N}\) from \(C\) to \(B\)
- C \(8 \times 10^{-5} \mathrm{~N}\) from \(C\) to \(A\)
- D \(8 \times 10^{-5} \mathrm{~N}\) from \(C\) to \(B\)
Answer & Solution
Correct Answer
(A) \(4 \times 10^{-5} \mathrm{~N}\) from \(C\) to \(A\)
Step-by-step Solution
Detailed explanation
Force, \(\quad F=\frac{k q^2}{d^2}=4 \times 10^{-5} \mathrm{~N}\) Now, \(A\) is touched by \(C\), then; Charge on \(C=q / 2\) Charge on \(A=q / 2\) So, force on \(C=\mathbf{F}_A+\mathbf{F}_B\)…
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