JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
A simple pendulum of mass \('m',\) length \('I'\) and charge \('+q'\) suspended in the electric field produced by two conducting parallel plates as shown. The value of deflection of pendulum in equilibrium position will be -
- A \(\tan ^{-1}\left[\frac{{q}}{{mg}} \times \frac{{C}_{2}\left({V}_{2}-{V}_{1}\right)}{\left({C}_{1}+{C}_{2}\right)({d}-{t})}\right]\)
- B \(\tan ^{-1}\left[\frac{{q}}{{mg}} \times \frac{{C}_{1}\left({V}_{1}+{V}_{2}\right)}{\left({C}_{1}+{C}_{2}\right)({d}-{t})}\right]\)
- C \(\tan ^{-1}\left[\frac{{q}}{m g} \times \frac{C_{1}\left(V_{2}-V_{1}\right)}{\left(C_{1}+C_{2}\right)(d-t)}\right]\)
- D \(\tan ^{-1}\left[\frac{{q}}{{mg}} \times \frac{{C}_{2}\left({V}_{1}+{V}_{2}\right)}{\left({C}_{1}+{C}_{2}\right)({d}-{t})}\right]\)
Answer & Solution
Correct Answer
(D) \(\tan ^{-1}\left[\frac{{q}}{{mg}} \times \frac{{C}_{2}\left({V}_{1}+{V}_{2}\right)}{\left({C}_{1}+{C}_{2}\right)({d}-{t})}\right]\)
Step-by-step Solution
Detailed explanation
Let \(E\) be electric field in air \(T \sin \theta=q E\) \(T \cos \theta=m g\) \(\tan \theta=\frac{q {E}}{m g}\) \({Q}=\left[\frac{{C}_{1} {C}_{2}}{{C}_{1}+{C}_{2}}\right]\left[{V}_{1}+{V}_{2}\right]\)…
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