Question

A large insulating thick sheet of thickness $2d$ carries a uniform charge per unit volume $\rho$. A particle of mass $m$, carrying a charge $q$ having a sign opposite to that of the sheet, is released from the surface of the sheet. The sheet does not offer any mechanical resistance to the motion of the particle. Find the oscillation frequency $\nu$ of the particle inside the sheet.

• A. $\nu =\frac{1}{2\pi }\sqrt{\frac{qp}{m{ϵ}_0}}$
• B. $\nu =\frac{1}{2\pi }\sqrt{\frac{2qp}{m{ϵ}_0}}$
• C. $\nu =\frac{1}{4\pi }\sqrt{\frac{qp}{m{ϵ}_0}}$
• D. $\nu =\frac{2}{2\pi }\sqrt{\frac{qp}{m{ϵ}_0}}$

Answer 1

The correct option is A $\nu =\frac{1}{2\pi }\sqrt{\frac{qp}{m{ϵ}_0}}$

If we draw a cuboid Gaussian surface of lateral cross section $A$ and height $h$ and width $2x$ where $x=$ distance from central line then $E$ will be pointing as shown (by symmetry)
${\varphi }_g=\oint E.ds=2Ea=\frac{q_{encl}}{{\in }_0}$ (Flux through lateral surface if $E=$ Electric field at distance $x$)
$q_{\text{enclosed}}=\left[\right(l\left)\right(2x\left)\right(h\left)\right]\rho$
$2EA=\frac{q_{\text{enclosed}}}{{ϵ}_0}=\frac{2Ax\rho }{{ϵ}_0}$
$=\frac{\left(2x\right)\left(l\right)h\rho }{{ϵ}_0}=\frac{2xA\rho }{{ϵ}_0}$
[as $lh=A$]
So, $E=-\frac{\rho }{{ϵ}_0}x$
Force $=qE=-\frac{q\rho }{{ϵ}_0}x$
$m\frac{d^{2}x}{d{t}^2}=\frac{-q\rho }{{ϵ}_0}x\Rightarrow {\omega }^2=\frac{q\rho }{{ϵ}_{0}m}$
$\omega =\sqrt{\frac{q\rho }{{ϵ}_{0}m}}$
$\nu =\frac{1}{2\pi }\sqrt{\frac{q\rho }{m{ϵ}_0}}$