Question

A parallel plate capacitor with air between its plates is charged using a battery and then disconnected from the battery.
$\begin{array}{cc}\text{Column-I}& \text{Column-II}\\ \text{A. Potential difference between the plates will}& \text{(p) Separation between the plates is increased by}\frac{K}{2}(K>2)\\ \text{decrease if}& \text{times the initial value and space between the plates}\\ & \text{after the separation has increased, is completely}\\ & \text{filled with a dielectric (here}K\text{is the dielectric constant}\\ \text{B. Electric field strength between the plates will reduce if}& \text{(q) Separation between the plates is increased}\\ \text{C. Electric energy stored in the capacitor will decrease if}& \text{(r)A dielectric with}K>1\text{is filled between plates of the capacitor}\\ \text{D. Electric energy density will decrease if}& \text{(s) Separation between the plates is reduced}\\ & \text{(t) None}\end{array}$

• A. $A-p,r;B-p,r;C-p,r;D-q,s$
• B. $A-p,r,s;B-p,r;C-p,r,s;D-p,r$
• C. $A-p,r;B-p;C-r,s;D-r$
• D. $A-p,q;B-r,s;C-s,t;D-p,q$

Answer 1

The correct option is B $A-p,r,s;B-p,r;C-p,r,s;D-p,r$

Since it is disconnected from the battery the charge on the capacitor should remain constant.

A) Potential difference $\Delta V=\frac{Q}{C}$
Generally for capacitor of thickness $d$, filled with air capacitance is $C_0=\frac{{ϵ}_{0}A}{d}$
for (p): $C=\frac{K{ϵ}_{o}A}{d+\frac{K}{2}d}=\frac{2K}{2+K}{C}_0>C_0(\because K>2)$
As $C$ increases $\Delta V$ reduces.
for(q): let $d_1$ be increment in separation $C=\frac{{ϵ}_{o}A}{d+d_1}$
As $C$ decreases $\Delta V$ increases.
for(r):$C=\frac{K{ϵ}_{o}A}{d}$
As $C$ increases $\Delta V$ reduces.
for(s): let $d_1$ be decrement in separation $C=\frac{{ϵ}_{o}A}{d-d_1}$
As $C$ increases $\Delta V$ reduces.
$A⟶p,r,s$

B) Electric field in a parallel plate capacitor filled with air is $E_0=\frac{Q}{{ϵ}_{0}A}$
for(p): $E=\frac{Q}{K{ϵ}_{0}A}=\frac{E_0}{K}$
$E$ reduces
for(q) As $E$ is independent of seperation between plates, remains same.
for (r):$E=\frac{Q}{K{ϵ}_{0}A}=\frac{E_0}{K}$
$E$ reduces
for(s) As $E$ is independent of seperation between plates, remains same.
$B⟶p,r$

C) Electric energy stored in the capacitor filled with air $U=\frac{Q^2}{2C_0}$
for(p): As $C$ increases $U$ reduces.
for(q): As $C$ decreases $U$ increases.
for(r): As $C$ increases $U$ reduces.
for(s): As $C$ increases $U$ reduces.
$C⟶p,r,s$

D)Electric energy density of the capacitor filled with air $E_{d_0}=\frac{U}{Ad}=\frac{Q^2}{2C_0}\frac{1}{Ad}=\frac{Q^2}{2{ϵ}_0{A}^2}$
for(p): $E_d=\frac{Q^2}{2K{ϵ}_0{A}^2}$
Electric energy density decreases.
for (q): As there is no dielectric , electric energy density remains same.
for(r): $E_d=\frac{Q^2}{2K{ϵ}_0{A}^2}$
Electric energy density decreases.
for (s): As there is no dielectric , electric energy density remains same.
$D⟶p,r$