Question

A potential difference of 50V is maintained between the two plates of a parallel plate condenser, separated by \[5mm\] of air. The energy stored in unit volume of air between the plates of condenser is-
(A). \[4.425\times {{10}^{-4}}J{{m}^{-3}}\]
(B). \[2.425\times {{10}^{-4}}J{{m}^{-3}}\]
(C). \[8.425\times {{10}^{-4}}J{{m}^{-3}}\]
(D) \[6.425\times {{10}^{-4}}J{{m}^{-3}}\]

Answer 1

Hint: Using the formula for capacitance of a parallel plate capacitor, calculate capacitance by substituting the corresponding values. Use the calculated value of capacitance and given value of potential difference to calculate energy stored in the condenser and divide by volume for energy stored per unit volume between the plates of the condenser.

Formula used:
  \[U=\dfrac{1}{2}C{{V}^{2}}\]
 \[C=\dfrac{{{\varepsilon }_{0}}A}{d}\]

Complete step by step solution:
A condenser, also known as a capacitor, is a device that stores electrical energy in the presence of an electrical field. Its potential to store charge is determined by capacitance. The SI unit of Capacitance is Farad ( \[F\] ).
The Energy stored between the plates of the capacitor is-
 \[U=\dfrac{1}{2}C{{V}^{2}}\] …………….. (1)
Here,
 \[U\] is the energy stored
 \[C\] is capacitance
 \[V\] is the potential difference applied across its ends.
The formula for calculating \[C\] is
 \[C=\dfrac{{{\varepsilon }_{0}}A}{d}\] …………………... (2)
Here,
 \[d\] is the distance between the plates
 \[A\] is the area of cross section of the capacitor
Given,
 \[\begin{align}
  & D=5mm \\
 & A=5\times 5m{{m}^{2}} \\
\end{align}\]
Substituting value in eq (2), we get,
 \[\begin{align}
  & C=\dfrac{8.85\times {{10}^{-12}}\times 5\times {{10}^{-3}}\times 5\times {{10}^{-3}}}{5\times {{10}^{-3}}} \\
 & \Rightarrow C=4.42\times {{10}^{-14}}F \\
 & \\
\end{align}\]
 Substituting the value of C from the above eq along with the given value of \[V\] in eq (1), we get,
 \[\begin{align}
  & U=\dfrac{1}{2}\times 4.42\times {{10}^{-14}}\times 50\times 50 \\
 & \Rightarrow U=5.52\times {{10}^{-11}}J \\
\end{align}\]
The total energy stored inside the capacitor is \[40\times {{10}^{-10}}J\] . Energy stored per unit volume is-
 \[\begin{align}
  & \dfrac{U}{V}=\dfrac{5.52\times {{10}^{-11}}}{{{(5\times {{10}^{-3}})}^{3}}} \\
 & \dfrac{U}{V}=4.4\times {{10}^{-4}}J{{m}^{-3}} \\
\end{align}\]
 Therefore the value of Energy per unit volume of the capacitor is \[4.4\times {{10}^{-4}}J{{m}^{-3}}\] .

So, the correct answer is “Option A”.

Note: A parallel plate capacitor holds energy by storing charge on its plates. The formulas for energy are
 \[U=\dfrac{1{{Q}^{2}}}{2C}\] , \[U=\dfrac{1}{2}QV\] . The other types of capacitors are spherical capacitors, cylindrical capacitors etc. The capacitance of capacitor can be changed by adding dielectric materials between its plates