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Question

(C) The time period of a simple pendulum is given by $T_0 = 2 \pi \sqrt{\frac{L}{g}}$.When the plates are charged,an electric field $E$ is created bet

Vedclass · Accepted Answer

$\left(\frac{g}{g+\frac{q E}{m}}\right)^{1 / 2}$

Vedclass · Answer

(C) The time period of a simple pendulum is given by $T_0 = 2 \pi \sqrt{\frac{L}{g}}$.When the plates are charged,an electric field $E$ is created between them. The force on the charged sphere due to the electric field is $F_e = qE$ acting downwards (assuming the charge $q$ is positive and the field is directed downwards).The effective acceleration due to gravity becomes $g_{	ext{eff}} = g + \frac{qE}{m}$.The new time period is $T = 2 \pi \sqrt{\frac{L}{g_{	ext{eff}}}} = 2 \pi \sqrt{\frac{L}{g + \frac{qE}{m}}}$.Therefore,the ratio is $\frac{T}{T_0} = \frac{2 \pi \sqrt{\frac{L}{g + \frac{qE}{m}}}}{2 \pi \sqrt{\frac{L}{g}}} = \sqrt{\frac{g}{g + \frac{qE}{m}}} = \left(\frac{g}{g + \frac{qE}{m}}ight)^{1/2}$.

$A$ small sphere of mass $m$ carrying a charge $q$ is hanging between two parallel plates by a string of length $L$ as shown in the figure. The time period of the pendulum is $T_0$. When the parallel plates are charged as shown,the time period changes to $T$. The ratio $T / T_0$ is equal to:

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