Question:
A parallel plate capacitor of capacitance
\[
12\,\mu\text{F}
\]
is charged to a potential of
\[
180\,\text{V}.
\]
If the distance between the plates of the capacitor is
\[
\frac{3}{\sqrt{\pi}}\ \text{mm},
\]
then the energy density of the electric field between the plates is
A parallel plate capacitor of capacitance
\[
12\,\mu\text{F}
\]
is charged to a potential of
\[
180\,\text{V}.
\]
If the distance between the plates of the capacitor is
\[
\frac{3}{\sqrt{\pi}}\ \text{mm},
\]
then the energy density of the electric field between the plates is
Show Hint
The energy density of an electric field depends only on the field strength:
\[
u=\frac12\varepsilon_0E^2.
\]
For a parallel plate capacitor,
\[
E=\frac{V}{d}.
\]
Notice that capacitance is not required once \(V\) and \(d\) are known.
Updated On: Jul 29, 2026
- \[ 25\times10^{-6}\ \text{Jm}^{-3} \]
- \[ 25\times10^{-3}\ \text{Jm}^{-3} \]
- \[ 50\times10^{-6}\ \text{Jm}^{-3} \]
- \[ 50\times10^{-3}\ \text{Jm}^{-3} \]
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
Energy density of an electric field is
\[
u=\frac12\varepsilon_0E^2,
\]
where
\[
E=\frac{V}{d}.
\]
Step 1: Calculate the electric field between the plates. Given, \[ V=180\ \text{V}, \] \[ d=\frac{3}{\sqrt{\pi}}\ \text{mm} = \frac{3\times10^{-3}}{\sqrt{\pi}}\ \text{m}. \] Hence, \[ E=\frac{V}{d} = \frac{180}{\frac{3\times10^{-3}}{\sqrt{\pi}}}. \] \[ E = 60\times10^{3}\sqrt{\pi}. \] \[ E = 6\times10^{4}\sqrt{\pi}\ \text{Vm}^{-1}. \]
Step 2: Substitute in the energy density formula. \[ u = \frac12\varepsilon_0E^2. \] Using \[ \varepsilon_0 = \frac{1}{36\pi\times10^{9}}, \] \[ u = \frac12 \left( \frac{1}{36\pi\times10^{9}} \right) \left( 6\times10^{4}\sqrt{\pi} \right)^2. \] \[ = \frac12 \left( \frac{1}{36\pi\times10^{9}} \right) \left( 36\times10^{8}\pi \right). \] \[ = \frac12\times10^{-1}. \] \[ = 5\times10^{-2}\ \text{Jm}^{-3}. \] \[ = 50\times10^{-3}\ \text{Jm}^{-3}. \] Therefore, \[ \boxed{ u=50\times10^{-3}\ \text{Jm}^{-3} } \] \[ \boxed{\text{Answer = (D)}} \]
Step 1: Calculate the electric field between the plates. Given, \[ V=180\ \text{V}, \] \[ d=\frac{3}{\sqrt{\pi}}\ \text{mm} = \frac{3\times10^{-3}}{\sqrt{\pi}}\ \text{m}. \] Hence, \[ E=\frac{V}{d} = \frac{180}{\frac{3\times10^{-3}}{\sqrt{\pi}}}. \] \[ E = 60\times10^{3}\sqrt{\pi}. \] \[ E = 6\times10^{4}\sqrt{\pi}\ \text{Vm}^{-1}. \]
Step 2: Substitute in the energy density formula. \[ u = \frac12\varepsilon_0E^2. \] Using \[ \varepsilon_0 = \frac{1}{36\pi\times10^{9}}, \] \[ u = \frac12 \left( \frac{1}{36\pi\times10^{9}} \right) \left( 6\times10^{4}\sqrt{\pi} \right)^2. \] \[ = \frac12 \left( \frac{1}{36\pi\times10^{9}} \right) \left( 36\times10^{8}\pi \right). \] \[ = \frac12\times10^{-1}. \] \[ = 5\times10^{-2}\ \text{Jm}^{-3}. \] \[ = 50\times10^{-3}\ \text{Jm}^{-3}. \] Therefore, \[ \boxed{ u=50\times10^{-3}\ \text{Jm}^{-3} } \] \[ \boxed{\text{Answer = (D)}} \]
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