Question

A parallel plate capacitor is made of square conducting plates of side $a$ and the seperation between the plates is $d$. The capacitor is connected with a battery of emf $V$ volts as shown in the figure. There is a dielectric slab of dimensions $a\times a\times d$ with dielectric constant $k$. At $t=0$, dielectric slab is given velocity $v_0$ towards the capacitor as shown in the figure. (Neglect the effect of gravity and electrostatic force acting on the dielectric slab outside of capacitor. Also ignore any type of frictional force acting on the dielectric during its motion). Let $x$ be the length of the dielectric inside the capacitor at $t=t\text{sec}.[l_0>>a]$

• A. Motion of dielectric slab is periodic but not simple harmonic motion.
• B. Motion of dielectric slab is simple harmonic motion.
• C. At any time, the slope of graph of total energy versus $x$ is twice the slope of graph of potential energy versus $x$.
• D. The value of maximum energy stored in the system is $\frac{1}{2}m{v}_0^2+\frac{{\epsilon }_0{a}^2{V}^2}{2d}(2k-1)$.

Answer 1

Answer: (C), (D)

The correct options are
C At any time, the slope of graph of total energy versus $x$ is twice the slope of graph of potential energy versus $x$.
D The value of maximum energy stored in the system is $\frac{1}{2}m{v}_0^2+\frac{{\epsilon }_0{a}^2{V}^2}{2d}(2k-1)$.

At time $t$, let $x$ be the length of the dielectric inside the capacitor.
Total capacitance
$C=\frac{{\epsilon }_0(a-x)a}{d}+\frac{{\epsilon }_{0}xak}{d}$ $=\frac{{\epsilon }_{0}a[a-x+kx]}{d}$ $=\frac{{\epsilon }_{0}a}{d}[a+(k-1\left)x\right]$

Potential energy of the system
$U=\frac{C{V}^2}{2}=\frac{{\epsilon }_{0}a{V}^2}{2d}[a+(k-1\left)x\right]$

Then, force $F$ acting on the dielectric slab
$F=\frac{dU}{dx}=\frac{{\epsilon }_{0}a{V}^2(k-1)}{2d}$

Force is constant, therefore the motion is not periodic or simple harmonic.

Acceleration of the slab $A=\frac{{\epsilon }_{0}a{V}^2(k-1)}{2md}$

When the slab starts entering the capacitor, velocity is $v_0$. At time $t$,

$v^2=v_0^2+2Ax=v_0^2+\frac{{\epsilon }_{0}a{V}^2(k-1)}{md}x$

Therefore, total energy at time $t$
$E=\frac{1}{2}m{v}^2+\frac{{\epsilon }_{0}a{V}^2}{2d}[a+(k-1\left)x\right]$
$=\frac{1}{2}m{v}_0^2+\frac{{\epsilon }_{0}a{V}^2(k-1)x}{2d}+\frac{{\epsilon }_{0}a{V}^2}{2d}[a+(k-1\left)x\right]$

$\therefore E_{max}=\frac{1}{2}m{v}_0^2+\frac{{\epsilon }_0{a}^2{V}^2(2k-1)}{2d}$ (when $x=a$)

Option D is correct.

Slope of total energy vs $x$ graph = $\frac{{\epsilon }_0{a}^2{V}^2}{2d}\left[2\right(k-1\left)\right]$ which is twice the slope of potential energy vs $x$ graph $\left(\frac{{\epsilon }_0{a}^2{V}^2}{2d}\right[k-1\left]\right)$
Option C is correct.