Two identical charged spheres are suspended f | vedclass

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

Question

(C) For equilibrium of one sphere,the forces acting are tension $T$,gravitational force $Mg$,and electrostatic force $F = \frac{K Q^2}{x^2}$.Resolving

Vedclass · Accepted Answer

$v \propto x^{-1/2}$

Vedclass · Answer

(C) For equilibrium of one sphere,the forces acting are tension $T$,gravitational force $Mg$,and electrostatic force $F = \frac{K Q^2}{x^2}$.Resolving the tension $T$ into components,we have $T \cos 	heta = Mg$ and $T \sin 	heta = F$.Dividing these equations,we get $	an 	heta = \frac{F}{Mg}$.Since $	heta$ is small,$	an 	heta \approx \sin 	heta = \frac{x/2}{l} = \frac{x}{2l}$.Thus,$\frac{x}{2l} = \frac{K Q^2}{x^2 Mg}$,which implies $Q^2 = \frac{Mg}{2Kl} x^3$,or $Q \propto x^{3/2}$.Differentiating both sides with respect to time $t$,we get $2Q \frac{dQ}{dt} = \frac{Mg}{2Kl} \cdot 3x^2 \frac{dx}{dt}$.Since $\frac{dQ}{dt}$ is constant,we have $Q \frac{dQ}{dt} \propto x^2 v$,where $v = \frac{dx}{dt}$.Substituting $Q \propto x^{3/2}$,we get $x^{3/2} \propto x^2 v$,which simplifies to $v \propto x^{3/2 - 2} = x^{-1/2}$.

Similar Questions

Four charges $+Q, -Q, +Q, -Q$ are placed at the corners of a square taken in order. At the centre of the square:

Three positive charges $q$ are placed at the vertices of an equilateral triangle. How do their electric field lines appear?

For the given figure,the direction of the electric field at point $A$ will be .........

Vedclass Test Series

Exam Paper Generator

Online Exam Module