Question

A parallel plate capacitor of area $A$ and separation $d$ is fully charged to potential difference $V$ and then removed from charging source. A dielectric slab of dielectric constant $K=2$, thickness $d$ and area $A/2$ is inserted as shown in figure. Let ${\sigma }_1$ be the free charge density at the conductor-dielectric surface, ${\sigma }_2$ be the charge density at the conductor- vacuum surface. The value of new capacitance and potential difference respectively are

• A. $\frac{3A{ϵ}_0}{2d},\frac{2}{3}V$
• B. $\frac{A{ϵ}_0}{2d},\frac{V}{2}$
• C. $\frac{3A{ϵ}_0}{d},\frac{V}{3}$
• D. $\frac{2A{ϵ}_0}{3d},\frac{4}{3}V$

Answer 1

The correct option is A $\frac{3A{ϵ}_0}{2d},\frac{2}{3}V$

Potential at each plate will remain the same over the entire area. If potential difference between them be $V^{\prime }$, then

$V^{\prime }=Ed$

Therefore, the electric field $\left(E\right)$ is also same inside the plates everywhere.
The given system can be understood as a combination of two capacitors in parallel.


So, equivalent capacitance will be

$C=C_1+C_2$

$\Rightarrow C=\frac{K\left(A/2\right){ϵ}_0}{d}+\frac{\left(A/2\right){ϵ}_0}{d}$

$\Rightarrow C=\frac{A{ϵ}_0}{d}+\frac{A{ϵ}_0}{2d}(\because K=2)$

$\therefore C=\frac{3A{ϵ}_0}{2d}$

After charging to $V\text{Volt}$ and disconnecting, the capacitor becomes isolated and charge on it remains same
i.e., $Q_i=C_{i}V$

$\Rightarrow Q_i=\left(\frac{A{ϵ}_0}{d}\right)\times V=\frac{A{ϵ}_{0}V}{d}$

Therefore,

$Q_i=Q_f$

$\Rightarrow C_{i}V=C_f{V}^{\prime }$

$\Rightarrow \frac{A{ϵ}_{0}V}{d}=\frac{3A{ϵ}_0}{2d}\times V^{\prime }$

$\therefore V^{\prime }=\frac{2V}{3}$

Hence, option $\left(a\right)$ is right.

Why this question? Tip: To maintain the conservation of charge, after battery is disconnected and dielectric inserted, the "potential difference will decrease since $C_{eq}$ has increased" Bottom line: In order to keep $\overrightarrow{E}$ same between the plates, the free charge density changes at the interface of conductor - dielectric & conductor - vacuum, due to redistribution of charge.