Question

A leaky parallel plate capacitor is filled with a material having dielectric constant $ K = 5 $ and electrical conductivity $ \sigma = 7.4 \times {19^{ - 12}}{\Omega ^{ - 1}}{m^{ - 1}} $. If the charge on the capacitor at the instant $ t = 0 $ is $ {q_0} = 8.5\mu C $, then calculate the leakage current at the instant $ t = 12s $.

Answer 1

Hint: To solve this question, we have to calculate the leaky current which is given by the formula, $ i = \dfrac{V}{R} $. Using the formula for the resistance and the voltage from the capacitance we will get the leaky current.

Formula used: The formulae used in solving this question are
 $\Rightarrow R = \dfrac{{\rho l}}{A} $, $ R $ is the resistance of a wire of length $ l $, cross sectional area $ A $, and having the resistivity $ \rho $.
 $\Rightarrow \rho = \dfrac{1}{\sigma } $, $ \rho $ is the resistivity, and $ \sigma $ is the conductivity.
 $\Rightarrow C = \dfrac{{{\varepsilon _0}a}}{d} $, $ C $ is the capacitance of a parallel plate capacitor, whose plates have a cross sectional area of $ a $ and are separated by a distance of $ d $.
 $\Rightarrow \tau = RC $, $ \tau $ is the time constant, $ C $ is the capacitance, and $ R $ is the resistance.

Complete step by step answer
A dielectric is an insulating material which restricts the flow of current through the capacitor. But it allows some of the current through it, which is called the leakage current.
Let the voltage applied across the capacitor be $ V $, and the resistance offered by the dielectric be $ R $. Then the leakage current can be given as
 $\Rightarrow i = \dfrac{V}{R} $.......................(1)
Now, the resistance is given by
 $\Rightarrow R = \dfrac{{\rho l}}{A} $
If the distance between the capacitor plates is $ d $ and their cross sectional area is $ a $, then the resistance is given by
 $\Rightarrow R = \dfrac{{\rho d}}{a} $.......................(2)
We know that the resistivity is equal to the inverse of the conductivity, that is,
 $\Rightarrow \rho = \dfrac{1}{\sigma } $
Substituting this in (2) we get
 $\Rightarrow R = \dfrac{d}{{\sigma a}} $.......................(3)
Substituting this in (1) we get
 $\Rightarrow i = \dfrac{{Va\sigma }}{d} $.......................(4)
Now, we now that the capacitance is given by
 $\Rightarrow C = \dfrac{{{\varepsilon _0}a}}{d} $
If the capacitor is fully filled with a dielectric of dielectric constant $ K $, then the capacitance becomes
 $\Rightarrow C' = kC $
 $ \Rightarrow C' = \dfrac{{K{\varepsilon _0}a}}{d} $
According to the question, $ K = 5 $. So we have
 $\Rightarrow C' = \dfrac{{5{\varepsilon _0}a}}{d} $.......................(5)
Now, the capacitance is related to the charge by
 $\Rightarrow q = C'V $
 $ \Rightarrow V = \dfrac{q}{{C'}} $
From (5)
 $\Rightarrow V = \dfrac{{qd}}{{5{\varepsilon _0}a}} $
Putting this in (4) we get
 $\Rightarrow i = \dfrac{{qda\sigma }}{{5{\varepsilon _0}ad}} $
On simplifying we get
 $\Rightarrow i = \dfrac{{q\sigma }}{{5{\varepsilon _0}}} $
Now, the charge at the time $ t = 0 $ is given as $ {q_0} $. Substituting this above, we get the initial current as
 $\Rightarrow {i_0} = \dfrac{{{q_0}\sigma }}{{5{\varepsilon _0}}} $.......................(6)
Now, we know that the leakage current as a function of time is given by
 $\Rightarrow i\left( t \right) = {i_o}{e^{ - t/\tau }} $
From (6)
 $\Rightarrow i\left( t \right) = \dfrac{{{q_0}\sigma }}{{5{\varepsilon _0}}}{e^{ - t/\tau }} $
The time constant of a capacitor is given by
 $\Rightarrow \tau = RC $
From (5) $ C = C' = \dfrac{{5{\varepsilon _0}a}}{d} $, therefore
 $\Rightarrow \tau = \dfrac{{5{\varepsilon _0}aR}}{d} $
Substituting (2)
 $\Rightarrow \tau = \dfrac{{5{\varepsilon _0}ad}}{{d\sigma a}} $
On simplifying, we get
 $\Rightarrow \tau = \dfrac{{5{\varepsilon _0}}}{\sigma } $
We know that $ {\varepsilon _0} = 8.85 \times {10^{ - 12}} $. Also, according to the question we have $ \sigma = 7.4 \times {19^{ - 12}}{\Omega ^{ - 1}}{m^{ - 1}} $. Substituting these above we get
 $\Rightarrow \tau = \dfrac{{5 \times 8.85 \times {{10}^{ - 12}}}}{{7.4 \times {{10}^{ - 12}}}} $
On solving we get
 $\Rightarrow \tau = 6s $
So from (7), the leakage current becomes
 $\Rightarrow i\left( t \right) = \dfrac{{{q_0}\sigma }}{{5{\varepsilon _0}}}{e^{ - t/6}} $
According to the question, $ {q_0} = 8.5\mu C $, $ \sigma = 7.4 \times {10^{ - 12}}{\Omega ^{ - 1}}{m^{ - 1}} $, and $ t = 12s $ as we know $ {\varepsilon _0} = 8.85 \times {10^{ - 12}} $. Substituting these above we get
 $\Rightarrow i\left( {12} \right) = \dfrac{{8.5 \times {{10}^{ - 6}} \times 7.4 \times {{10}^{ - 12}}}}{{5 \times 8.85 \times {{10}^{ - 12}}}}{e^{ - 12/6}}\mu A $
On solving we get
 $\Rightarrow i\left( {12} \right) = 0.192\mu A $
Hence, the leakage current at the instant $ t = 12s $ is equal to $ 0.192\mu A $.

Note
We should not forget to multiply the value of the dielectric constant with the original value of the capacitance while calculating the time constant. Inside a dielectric medium, the permittivity is different than that of the free space.