As shown in the figure,a configuration of two | vedclass

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

Question

(B) For the upper charge to be in equilibrium along the inclined plane,the component of gravitational force acting down the plane must be balanced by

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(B) For the upper charge to be in equilibrium along the inclined plane,the component of gravitational force acting down the plane must be balanced by the electrostatic repulsive force acting up the plane.Let $r$ be the distance between the two charges. From the geometry,$h = r \sin 30^{\circ}$,so $r = \frac{h}{\sin 30^{\circ}} = 2h$.The force balance equation is: $mg \sin 30^{\circ} = \frac{1}{4 \pi \varepsilon_0} \frac{q_0^2}{r^2}$.Substituting $r = 2h$ and the given values $(m = 0.02 \, kg, g = 10 \, ms^{-2}, q_0 = 2 	imes 10^{-6} \, C, \sin 30^{\circ} = 0.5)$:$0.02 	imes 10 	imes 0.5 = 9 	imes 10^9 	imes \frac{(2 	imes 10^{-6})^2}{(2h)^2}$$0.1 = 9 	imes 10^9 	imes \frac{4 	imes 10^{-12}}{4h^2}$$0.1 = \frac{9 	imes 10^{-3}}{h^2}$$h^2 = \frac{9 	imes 10^{-3}}{0.1} = 9 	imes 10^{-2} \, m^2$$h = 0.3 \, m = 300 	imes 10^{-3} \, m$.Comparing this with $h = x 	imes 10^{-3} \, m$,we get $x = 300$.

Similar Questions

Two charges are placed as shown in the figure. Where should a third charge be placed so that it remains at rest?

Three charges $4q, Q$ and $q$ are in a straight line along the $x$-axis at the positions $x=0 \ m$,$x=\frac{1}{2} \ m$,and $x=1 \ m$ respectively. The resultant force on $q$ will be zero if $Q$ is equal to:

$A$ charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to:

The law governing the force between electric charges is known as

Vedclass Test Series

Exam Paper Generator

Online Exam Module