Question

Consider the situation shown in figure.


The plates of the capacitor have plate area $A$ and are clamped in the laboratory. The dielectirc slab is released from rest with a length $a$ inside the capacitor. Neglecting any effect of friction or gravity, show that the slab will execute periodic motion and find its time period.

• A. $\sqrt{\frac{mld(l-a)}{{\epsilon }_{0}A{E}^{2}A(K-1)}}$
• B. $2\sqrt{\frac{mld(l-a)}{{\epsilon }_{0}A{E}^{2}A(K-1)}}$
• C. $4\sqrt{\frac{mld(l-a)}{{\epsilon }_{0}A{E}^{2}A(K-1)}}$
• D. $8\sqrt{\frac{mld(l-a)}{{\epsilon }_{0}A{E}^{2}A(K-1)}}$

Answer 1

The correct option is D $8\sqrt{\frac{mld(l-a)}{{\epsilon }_{0}A{E}^{2}A(K-1)}}$

Capacitance of the portion with dielectrics,
$C_1=\frac{aK{\epsilon }_{0}A}{ld}$

Capacitance of the portion without dielectrics,
$C_2=\frac{{\epsilon }_0(l-a)A}{ld}$


$\therefore$Net capacitance $C=C_1+C_2=\frac{{\epsilon }_{0}A}{ld}[Ka+(l-a\left)\right]$

$\Rightarrow C=\frac{\epsilon A}{ld}[l+a(k-1\left)\right]$

Consider the motion of dielectric in the capacitor.
Let, it further moves a distance $dx$,which causes an increase of capacitance by $dC$.

$\therefore$ Charge flowing, $dQ=\left(dC\right)E$

The work done by the battery $dW=VdQ=E\left(dC\right)E=E^{2}dC$

Let force acting on it be $F$.

$\therefore$Work done by the force during the displacement, $dW=Fdx$

$\therefore$ Increase in energy stored in the capacitor
$\frac{1}{2}\left(dC\right)E^2=Fdx$

$\Rightarrow F=\frac{1}{2}\frac{E^{2}dC}{dx}$

As, $C=\frac{{\epsilon }_{0}A}{ld}[l+a(K-1\left)\right]$ (here $x=a$)

$\Rightarrow \frac{dC}{da}=\frac{d}{da}\left[\frac{{\epsilon }_{0}A}{ld}\{l+a(K-1\left)\right\}\right]=\frac{{\epsilon }_{0}A}{ld}(K-1)$

Now,
$F=\frac{1}{2}{E}^2\frac{dC}{da}=\frac{1}{2}{E}^2\left[\frac{{\epsilon }_{0}A}{ld}\right(K-1\left)\right]$

So, the acceleration of the dielectric, $a_d=\frac{F}{m}=\frac{E^2{\epsilon }_{0}A(K-1)}{2ldm}$

If time required to move $(l-a)$ distance is $t$ then from the equation of motion, $(l-a)=\frac{1}{2}{a}_d{t}^2$

$\Rightarrow t=\sqrt{\frac{2(l-a)}{a_d}}=\sqrt{\frac{2(l-a)2ldm}{E^2{\epsilon }_{0}A(K-1)}}$

$\Rightarrow t=\sqrt{\frac{4mld(l-a)}{{\epsilon }_{0}A{E}^{2}A(K-1)}}$

For the complete cycle time period will be four times of the time required to move $(l-a)$ distance.

$\therefore$Time period$=4t=8\sqrt{\frac{mld(l-a)}{{\epsilon }_{0}A{E}^{2}A(K-1)}}$

Hence, option (d) is the correct answer.