Capacitor

Energy density in air gap $=\frac{1}{2} \varepsilon_0 E^2=u_E$

Energy density of dielectric $=\frac{1}{2} \varepsilon_0 K \frac{E^2}{K^2}=\frac{u_E}{K}$

Now, energy stored $=$ energy density × volume

∴ Total energy stored

$=\frac{3}{4} \cdot x \cdot \frac{u_E}{K}+\frac{1}{4} \cdot x \cdot u_E[x=$ volume of capacitor $]$

$=x u_E\left(\frac{3}{4 K}+\frac{1}{4}\right) \therefore K=3$;

$=x u_E\left(\frac{2}{4}\right)=\frac{x u_E}{2}$

Energy stored in dielectric $=\frac{3}{4} \cdot x \cdot \frac{u_E}{K}=\frac{x u_E}{4}$

So, \% of energy stored in the dielectric

$ =\frac{\frac{x u_E}{4}}{\frac{x u_E}{2}} \times 100=50 \% $

Given the radius of the capacitor plates $ R = 0.1 \, \mathrm{m} $, we can calculate the area of the plates as:

$ A = \pi R^2 = 3.14 \times (0.1)^2 \, \mathrm{m}^2 $

We are also given that the rate of change of the electric field between the capacitor plates is:

$ \frac{dE}{dt} = 6 \times 10^{13} \, \frac{\mathrm{V}}{\mathrm{m} \cdot \mathrm{s}} $

The displacement current $ I_d $ between the plates is calculated using the formula:

$ I_d = \varepsilon_0 \frac{d\phi}{dt} $

where $ \frac{d\phi}{dt} = A \frac{dE}{dt} $. Hence, the formula becomes:

$ I_d = \varepsilon_0 A \left(\frac{dE}{dt}\right) $

Substituting the values, where $ \varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{F/m} $, we have:

$ I_d = 8.85 \times 10^{-12} \times 3.14 \times (0.1)^2 \times 6 \times 10^{13} $

Calculating the above expression yields:

$ I_d = 16.67 \, \mathrm{A} $

Therefore, the value of the displacement current is $ 16.67 \, \mathrm{A} $.

In a uniform electric field, potential difference across two parallel plates of capacitor,

$V = Ed$

where, E is electric field and d is separation between plates.

$ \Rightarrow V = 2000 \times (1 \times {10^{ - 3}}) = 2V$

Since capacitors C₁ and C₂ are parallel, so their equivalent capacitance will be (C₁ + C₂). Now, (C₁ + C₂) and C₃ are in series, so the net equivalent capacitance of circuit will be,

${1 \over C} = {1 \over {{C_3}}} + {1 \over {{C_1} + {C_2}}} = {{{C_1} + {C_2} + {C_3}} \over {({C_1} + {C_2}){C_3}}}$

or, $C = {{({C_1} + {C_2}){C_3}} \over {{C_1} + {C_2} + {C_3}}}$

Since V is the voltage of the battery, so charge on this system,

$q = CV \Rightarrow q = {{({C_1} + {C_2}){C_3}V} \over {({C_1} + {C_2} + {C_3})}}$

If the capacitance C₃ breaks down, then total equivalent capacitance will become

$C' = {C_1} + {C_2}$

$\therefore$ New charge stored,

$q' = C'V$

$ \Rightarrow q' = ({C_1} + {C_2})V$

Now, change in total charge,

$\Delta q = q' - q$ [$\because$ $q' > q$]

$ = ({C_1} + {C_2})V - {{({C_1} + {C_2}){C_3}V} \over {{C_1} + {C_2} + {C_2}}}$

or, $\Delta q = ({C_1} + {C_2})V\left[ {1 - \left( {{{{C_3}} \over {{C_1} + {C_2} + {C_3}}}} \right)} \right]$