A small sphere carrying a charge q is suspend | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

What is the gametophytic stage of Pteridophytes?

In the figure shown, if the internal resistance of the battery is $1\, \Omega$, the reading of the ammeter will be ............... $A$.

When an ideal diatomic gas is heated at constant pressure,the fraction of the heat energy supplied which increases the internal energy of the gas is:

Which of the following is an example of a biofertilizer?

$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]$ is equal to :

Vedclass Test Series

Exam Paper Generator

Online Exam Module