Question 45: Electrostats & Capacitance

Question

A bob of a simple pendulum has a mass of 4 g and a charge of $20 \mu \mathrm{C}$. If it is at rest in a uniform horizontal electric field of intensity $1000 \mathrm{Vm}^{-1}$, the angle that the pendulum makes with the vertical at equilibrium is:

Accepted Answer

Let the pendulum bob of mass $m=4\,$g $=0.004\,$kg carry a charge $q=20	imes10^{-6}\,$C in a horizontal field $E=1000\,$V/m. At equilibrium the horizontal electric force balances the component of tension:
Electric force:
$$F_{e}=qE=(20	imes10^{-6}\,	ext{C})(1000\,	ext{V/m})=0.02\,	ext{N}.$$
Weight:
$$W=mg=(0.004\,	ext{kg})(9.8\,	ext{m/s}^2)=0.0392\,	ext{N}.$$
At angle $	heta$, equilibrium requires
$$	an	heta=\frac{F_{e}}{W}=\frac{0.02}{0.0392}\approx0.51\approx\frac12.$$
Hence
$$	heta=	an^{-1}\!\frac12,$$
which is Option A.

Answer

$	an ^{-1} \frac{1}{2}$

Answer

$60^{\circ}$

Answer

$30^{\circ}$

Answer

$	an ^{-1} 2$

Electrostats & Capacitance

A bob of a simple pendulum has a mass of 4 g and a charge of 20 \mu \mathrm{C}. If it is at rest in a uniform horizontal electric field of intensity 1000 \mathrm{Vm}^{-1}, the angle that the pendulum makes with the vertical at equilibrium is:

Choose the correct answer: