JEE Mains · Physics · STD 12 - 1. Electric charges and fields
A simple pendulum has a bob with mass m and charge q. The pendulum string has negligible mass. When a uniform and horizontal electric field \( \vec{E} \) is applied, the final tension in the string changes.The final tension in the string, when pendulum attains an equilibrium position is __________ .
(g: acceleration due to gravity)
- A \( mg-qE \)
- B \( mg+qE \)
- C \( \sqrt{m^{2}g^{2}+q^{2}E^{2}} \)
- D \( \sqrt{m^{2}g^{2}-q^{2}E^{2}} \)
Answer & Solution
Correct Answer
(C) \( \sqrt{m^{2}g^{2}+q^{2}E^{2}} \)
Step-by-step Solution
Detailed explanation
\( T=\sqrt{(qE)^{2}+(mg)^{2}} \)
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