Question

A parallel plate capacitor of capacitance 1 μF is charged to a potential difference of 20 V. The distance between plates is 1 μm. The energy density between the plates of the capacitor is:

• A. \( 2 \times 10^{-4} \, {J/m}^3 \)
• B. \( 1.8 \times 10^5 \, {J/m}^3 \)
• C. \( 1.8 \times 10^3 \, {J/m}^3 \)
• D. \( 2 \times 10^2 \, {J/m}^3 \)

Hint:

For energy density in capacitors, remember to use the formula \( U = \frac{1}{2} \epsilon_0 E^2 \) and calculate the electric field \( E \) first.

Answer 1

The Correct Option is C

Given: \[C = 1 \, \mu{F}, \quad V = 20 \, {V}, \quad d = 1 \, \mu{m}\] The energy density is calculated using: \[U = \frac{1}{2} \epsilon_0 E^2\] The electric field is determined by: \[E = \frac{V}{d} = \frac{20 \times 10^6}{1 \times 10^{-6}} = 20 \times 10^6 \, {V/m}\] The resulting energy density is: \[U = \frac{1}{2} \epsilon_0 E^2 = 1.77 \times 10^3 \, {J/m}^3\]