Two identical charged particles,each having a | vedclass

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

Question

(A) For the particles to be in limiting equilibrium,the electrostatic repulsive force must be equal to the maximum static frictional force.Electrostat

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) For the particles to be in limiting equilibrium,the electrostatic repulsive force must be equal to the maximum static frictional force.Electrostatic force $F_e = \frac{kq^2}{L^2}$,where $k = 9 	imes 10^9 \, N \cdot m^2/C^2$.Frictional force $F_f = \mu mg$,where $\mu = 0.25$,$m = 10 	imes 10^{-3} \, kg$,and $g = 10 \, m/s^2$.Equating the two: $\frac{kq^2}{L^2} = \mu mg$.$L^2 = \frac{kq^2}{\mu mg} = \frac{(9 	imes 10^9) 	imes (2.0 	imes 10^{-7})^2}{0.25 	imes (10 	imes 10^{-3}) 	imes 10}$.$L^2 = \frac{9 	imes 10^9 	imes 4 	imes 10^{-14}}{0.25 	imes 0.1} = \frac{36 	imes 10^{-5}}{0.025} = \frac{36 	imes 10^{-5}}{25 	imes 10^{-3}} = 1.44 	imes 10^{-2} \, m^2$.$L = \sqrt{1.44 	imes 10^{-2}} = 1.2 	imes 10^{-1} \, m = 0.12 \, m = 12 \, cm$.

Similar Questions

The electric field due to a uniformly charged sphere of radius $R$ as a function of the distance from its center is represented by which of the following graphs?

$A$ charge $+q$ is situated at a distance $d$ away from both the sides of a grounded conducting $L$-shaped sheet as shown in the figure. The force acting on the charge $+q$ is:

Two identical small spheres carry charges $Q_1$ and $Q_2$ $(Q_1 >> Q_2)$. The force between them is $F_1$. The spheres are brought into contact and then placed at the same distance. The new force between them is $F_2$. Then $F_1/F_2$ will be:

In the given figure,two tiny conducting balls of identical mass $m$ and identical charge $q$ hang from non-conducting threads of equal length $L$. Assume that $\theta$ is so small that $\tan \theta \approx \sin \theta$,then for equilibrium,$x$ is equal to:

Two point charges $Q$ and $-Q/4$ are separated by a distance $x$. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module