Question

The line AA' is on a changed infinite conducting plane which is perpendicular to the plane of the paper. The plane has a surface density of charge $\sigma$ and B is a ball of mass m with a like charge of magnitude q. B is connected by a string from a point on the line AA'. The tangent of the angle $\left(\theta \right)$ formed between the line AA' and the string is:

• A. $\frac{q\sigma }{2{ϵ}_{0}mg}$
• B. $\frac{q\sigma }{4\pi {ϵ}_{0}mg}$
• C. $\frac{q\sigma }{2\pi {ϵ}_{0}mg}$
• D. $\frac{q\sigma }{{ϵ}_{0}mg}$

Answer 1

The correct option is D $\frac{q\sigma }{{ϵ}_{0}mg}$

IN THE GIVEN FIGURE,

$Fe$ = FORCE PRODUCED BY INFINITELY CHARGED SHEET

$T$ = TENSION PRESENT IN THE STRING

$T\cos \theta$ AND $T\sin \theta$ ARE THE RECTANGULAR COMPONENTS OF THE TENSION.

ACCORDING TO THE SITUATION,

GRAVITATIONAL FORCE = $T\cos \theta$

$\therefore$ $mg$ = $T\cos \theta$

$T=\frac{mg}{cos\theta }$ ...$\left(i\right)$

NOW BALANCING THE HORIZONTAL COMPONENT OF FORCE ON THE PARTICLE.

FORCE DUE TO ELECTRIC FIELD = $T\sin \theta$

$Fe$ = $T\sin \theta$

FROM $\left(i\right)$,

$Fe=\frac{tan\theta }{mg}$

AND FROM THE CONCEPT OF INFINITLY CHARGED SHEET,

$Fe=E\ast q$ ...$\left(ii\right)$

$E=\frac{\sigma }{{ϵ}_0}$ ...$\left(iii\right)$

FROM EQUATIONS $\left(ii\right)$ AND $\left(iii\right)$,

$tan0=\frac{q\sigma }{{ϵ}_{0}mg}$

HENCE THE OPTION (D) IS CORRECT