A simple pendulum of mass m, length l, and ch | vedclass

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(D) Let $E$ be the electric field in the air region between the plates.In the equilibrium position, the forces acting on the pendulum are the tension

Vedclass · Accepted Answer

$	an^{-1}\left[\frac{q}{mg} 	imes \frac{C_{2}(V_{1}+V_{2})}{(C_{1}+C_{2})(d-t)}ight]$

Vedclass · Answer

(D) Let $E$ be the electric field in the air region between the plates.In the equilibrium position, the forces acting on the pendulum are the tension $T$, the gravitational force $mg$, and the electric force $qE$.Resolving the forces, we have:$T \sin 	heta = qE$$T \cos 	heta = mg$Dividing these equations, we get $	an 	heta = \frac{qE}{mg}$.The system acts as two capacitors $C_{1}$ (air) and $C_{2}$ (dielectric) in series.The potential difference across the plates is $V = V_{1} + V_{2}$.The charge $Q$ on the series combination is $Q = \left[\frac{C_{1}C_{2}}{C_{1}+C_{2}}ight](V_{1}+V_{2})$.The electric field $E$ in the air region is given by $E = \frac{Q}{A\epsilon_{0}}$.Substituting $Q$, we get $E = \left[\frac{C_{1}C_{2}}{C_{1}+C_{2}}ight] \frac{(V_{1}+V_{2})}{A\epsilon_{0}}$.Since $C_{1} = \frac{\epsilon_{0}A}{d-t}$, we have $\frac{1}{A\epsilon_{0}} = \frac{1}{C_{1}(d-t)}$.Substituting this into the expression for $E$, we get $E = \frac{C_{2}(V_{1}+V_{2})}{(C_{1}+C_{2})(d-t)}$.Thus, the deflection angle is $	heta = 	an^{-1}\left[\frac{q}{mg} 	imes \frac{C_{2}(V_{1}+V_{2})}{(C_{1}+C_{2})(d-t)}ight]$.

$A$ simple pendulum of mass $m$, length $l$, and charge $+q$ is suspended in the electric field produced by two conducting parallel plates as shown. The value of the deflection of the pendulum in the equilibrium position will be -

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