A small sphere carrying a charge q is hanging | vedclass

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The initial time period of the pendulum is $T_0 = 2\pi \sqrt{\frac{L}{g}}$.When the plates are charged,an electric field $E$ is created. The force on the charge $q$ is $F_e = qE$ acting downwards.The net downward force on the sphere is $F_{net} = mg + qE$.The effective acceleration $g'$ is given by $g' = \frac{F_{net}}{m} = g + \frac{qE}{m}$.The new time period is $T = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{g + \frac{qE}{m}}}$.Taking the ratio $T/T_0$,we get:$\frac{T}{T_0} = \frac{2\pi \sqrt{\frac{L}{g + \frac{qE}{m}}}}{2\pi \sqrt{\frac{L}{g}}} = \sqrt{\frac{g}{g + \frac{qE}{m}}} = \left( \frac{g}{g + \frac{qE}{m}} \right)^{1/2}$.Thus,the correct option is $(c)$.

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

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