Question

A parallel plate capacitor of area $A$ and separation $d$ is charged to potential difference $V$ and removed from the charging source. A dielectric slab of constant $K=5$, thickness $d$ and area $\frac{A}{3}$ is inserted, as shown in the figure. Let ${\sigma }_1$ be free charge density at the conductor-dielectric surface and ${\sigma }_2$ be the charge density at the conductor-vacuum surface :

• A. The electric field have the same value inside the dielectric as in the free space between the plates.
• B. The ratio $\frac{{\sigma }_1}{{\sigma }_2}$ is equal to $\frac{1}{5}$
• C. The new capacitance is $\frac{7{ϵ}_{0}A}{3d}$
• D. The new potential difference is $\frac{3}{7}V$

Answer 1

Answer: (A), (C), (D)

The new capacitance is $\frac{7{ϵ}_{0}A}{3d}$
The electric field have the same value inside the dielectric as in the free space between the plates.
The new potential difference is $\frac{3}{7}V$

Since the capacitors are attached in parallel, the potential across the capacitors is same and so is the electric field inside the capacitors $(Ed=V)$.Therefore,

$letC=\frac{A{ϵ}_0}{d}$

Now,

$\frac{q}{\frac{\left(2A{ϵ}_0\right)}{\left(3d\right)}}=\frac{(CV-q)}{\frac{(A\kappa {ϵ}_0}{\left(3d\right)})}$

$\frac{3q}{2C}=\frac{3(CV-q)}{5C}\Rightarrow 7q=2CV$

Therefore $V=\frac{q}{\frac{2A{ϵ}_0}{\left(3d\right)}}=\frac{\frac{2CV}{7}}{\frac{2C}{3}}=\frac{3V}{7}$

Also $C_{net}=\frac{2A{ϵ}_0}{\left(3d\right)}+\frac{A\kappa {ϵ}_0}{\left(3d\right)}=\frac{7A{ϵ}_0}{\left(3d\right)}$

Also ratio of charge densities,

$\frac{{\sigma }_2}{{\sigma }_1}=\frac{\frac{q}{\frac{2A}{3}}}{\frac{CV-q)}{\frac{A}{3}}}=\frac{\frac{\frac{2CV}{7}}{\frac{2A}{3}}}{\frac{\frac{5CV}{7}}{\frac{A}{3}}}=\frac{1}{5}$