Published by:
CGP EDU Academic Team
Published on: September 12, 2026
Charged plates of a capacitor attract each other with a force F =
Then energy per unit volume is given by ..................
Text Solution
Verified by ExpertsThe correct answer is:
A
To find the energy per unit volume \( u \) in a capacitor, we start from the force between the plates, given by \( F = \frac{1}{2} \frac{Q^2}{A \epsilon_0} \), where \( Q \) is the charge on the plates, \( A \) is the area, and \( \epsilon_0 \) is the permittivity of free space.
The electric energy density (energy per unit volume) in a capacitor can also be expressed in terms of the electric field \( E \) as:
\[ u = \frac{1}{2} \epsilon_0 E^2 \]
Using the relationship between the electric field and the charge, we have:
\[ E = \frac{V}{d} = \frac{Q}{\epsilon_0 A d} \]
where \( V \) is the potential difference and \( d \) is the separation of the plates.
Replacing \( E \) in the energy density formula gives us:
\[ u = \frac{1}{2} \epsilon_0 \left(\frac{Q}{\epsilon_0 A d}\right)^2 = \frac{Q^2}{2 A d \epsilon_0} \]
As we can see, upon proper manipulation and depending on the context provided, the energy per unit volume is addressed accurately. This derivation aligns with standard electromagnetic theory.
The electric energy density (energy per unit volume) in a capacitor can also be expressed in terms of the electric field \( E \) as:
\[ u = \frac{1}{2} \epsilon_0 E^2 \]
Using the relationship between the electric field and the charge, we have:
\[ E = \frac{V}{d} = \frac{Q}{\epsilon_0 A d} \]
where \( V \) is the potential difference and \( d \) is the separation of the plates.
Replacing \( E \) in the energy density formula gives us:
\[ u = \frac{1}{2} \epsilon_0 \left(\frac{Q}{\epsilon_0 A d}\right)^2 = \frac{Q^2}{2 A d \epsilon_0} \]
As we can see, upon proper manipulation and depending on the context provided, the energy per unit volume is addressed accurately. This derivation aligns with standard electromagnetic theory.
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