A simple pendulum of mass 'm', length 'ℓ' 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 :

Engineering

Physics

Parallel Plate Capacitor

Question

$\tan^{- 1} [\frac{q}{mg} \times \frac{C_{2} (V_{2} - V_{1})}{(C_{1} + C_{2}) (d - t)}]$

$\tan^{- 1} [\frac{q}{mg} \times \frac{C_{2} (V_{1} + V_{2})}{(C_{1} + C_{2}) (d - t)}]$

$\tan^{- 1} [\frac{q}{mg} \times \frac{C_{1} (V_{2} + V_{2})}{(C_{1} + C_{2}) (d - t)}]$

$\tan^{- 1} [\frac{q}{mg} \times \frac{C_{1} (V_{2} - V_{1})}{(C_{1} + C_{2}) (d - t)}]$

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Considering one part of capacitor with dielectric as C₂ and remaining with pendulum as C₁ (As is not clearly mentioned in question)

Charge on each capacitor

$= [V_{2} - (- V_{1})] \times [\frac{C_{1} C_{2}}{C_{1} + C_{2}}]$

$= \frac{C_{1} C_{2}}{C_{1} + C_{2}} (V_{1} + V_{2})$

Potential difference across C₁

$V_{1} = \frac{Q}{C_{1}} = \frac{C_{2} (V_{1} + V_{2})}{(C_{1} + C_{2})}$

$E = \frac{C_{2} (V_{1} + V_{2})}{(C_{1} + C_{2}) (d - t)}$

Force $= Eq = \frac{{qC}_{2}}{(C_{1} + C_{2})} \frac{(V_{1} + V_{2})}{(d - t)}$

$θ = \tan^{- 1} [\frac{q}{mg} \times \frac{C_{2} (V_{1} + V_{2})}{(C_{2} + C_{2}) (d - t)}]$

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