Question

A parallel plate capacitor contains a mica sheet (thickness\[{10^{ - 3}}m\]) and a sheet of fiber (thickness\[0.5 \times {10^{ - 3}}m\]). The dielectric constant of mica is 8 and that of the fiber is \[2.5\]. Assuming that the fiber breaks down when subjected to an electric field of \[6.4 \times {10^6}V{m^{ - 1}}\], find the maximum safe voltage (in kV) that can be applied to the capacitor.
\[(A)2kV\]
\[(B)3kV\]
\[(C)5kV\]
\[(D)7kV\]

Answer 1

Hint: From the given data, first we have to find the total electric field due to both the plates of the capacitor. Use it to find the potential difference between them. Calculate the maximum safe voltage that can be applied to the capacitor. Convert the final answer from Volt to kilovolt.

Formula used:
There are two capacitors and they are connected in series. If \[{V_1}\] be the potential across mica capacitor and \[{V_2}\] for fiber capacitor, then
\[V = {V_1} + {V_2} = {E_1}{d_1} + {E_1}{d_2}\]
From the given data,
Electric field \[ = {E_1} = {E_2} = 6.4 \times {10^6}V{m^{ - 1}}\]
Thickness \[{d_1} = {10^{ - 3}}m,{d_2} = 0.5 \times {10^{ - 3}}m\]
Dielectric constant \[{K_1} = 8,{K_2} = 2.5\]

Complete step-by-step solution:
The potential difference across the plates of the capacitor
\[V = {V_1} + {V_2}\]
Where,
\[{V_1}\] - The potential across mica capacitor and \[{V_2}\]- The potential across fiber capacitor
As the sheet of fiber breaks down when it is subjected to an electric field of \[6.4 \times {10^6}V/m\].
\[V = {E_1}{d_1} + {E_2}{d_2} = \dfrac{{{E_1}{d_1}}}{{{K_1}}} + \dfrac{{{E_2}{d_2}}}{{{K_2}}}\]
\[ \Rightarrow V = \dfrac{{6.4 \times {{10}^6} \times {{10}^{ - 3}}}}{8} + \dfrac{{6.4 \times {{10}^6} \times 0.5 \times {{10}^{ - 3}}}}{{2.5}}\]
\[ \Rightarrow V = 2.08 \times {10^3}V = 2.08kV.\]
The maximum safe voltage (in kV) that can be applied to the capacitor is 2kV.
Hence, option A is correct.

Additional information:
The charge of a capacitor is found by calculating the product of voltage across the capacitor and capacitance. When we use a dielectric between the plates the capacitance will increase. The total capacitance varies according to the parallel and series connection of the capacitors.

Note:For a parallel plate capacitor charge density is opposite for both the plates. And the electric field between the plates is directed from positive to the negatively charged plate. This electric field will give the potential difference value between the plates. If both plates carry the same charge (either positive or both negative), then the net electric field between them will be zero.