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(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.Initially,when the capacitor is uncharged,the effective accel

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The time period of a simple pendulum is given by $T = 2\pi \sqrt{\frac{L}{g_{eff}}}$.Initially,when the capacitor is uncharged,the effective acceleration due to gravity is $g_{eff} = g$. Thus,$T_0 = 2\pi \sqrt{\frac{L}{g}}$.When the capacitor is charged,an electric field $E$ is produced between the plates. The force on the charge $q$ is $F = qE$ acting downwards (assuming the charge is positive and the field is directed downwards).The effective acceleration becomes $g_{eff} = g + \frac{qE}{m}$.Therefore,the new time period is $T = 2\pi \sqrt{\frac{L}{g + \frac{qE}{m}}}$.The ratio $T/T_0$ is $\frac{2\pi \sqrt{\frac{L}{g + \frac{qE}{m}}}}{2\pi \sqrt{\frac{L}{g}}} = \sqrt{\frac{g}{g + \frac{qE}{m}}} = {\left( {\frac{g}{{g + \frac{{qE}}{m}}}} \right)^{1/2}}$.

$A$ small sphere carrying a charge $q$ is suspended by a string of length $L$ between two parallel plates. The time period of the pendulum is $T_0$. When the capacitor is charged,the time period changes to $T$. The ratio $T/T_0$ is equal to:

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