Question

The distance between plates of a parallel plate capacitor is $5d$. Let the positively charged plate is at $x=0$ and negatively charged plate is at $x=5d$. Two slabs, one of conductor and the other of a dielectric of equal thickness $d$ are inserted between the plates as shown in figure. Potential versus distance graph will look like:

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Electric field between two plates of capacitor is given by,
$\frac{\sigma }{K{ϵ}_0}$

When $K=1$ then $E=\frac{\sigma }{{ϵ}_0}$

For $K=K$ then $E=\frac{\sigma }{K{ϵ}_0}$

When $K=\infty$ then $E=0$. From the formula $V=E.d$.

Now positive plate at $x=0$ is at higher potential and potential drops linearly as $E$ is constant.

But as $E$ is the slope of potential versus distance curve hence inside the dielectric as $E$ decreases hence slope of $v$ versus $x$ curve for the interval $x=3d$ to $x=4d$ also decreases.