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

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

Answer 2

Step 1: Given data.
Capacitance, \( C = 1 \, \mu F = 1 \times 10^{-6} \, F \)
Potential difference, \( V = 20 \, V \)
Distance between plates, \( d = 1 \, \mu m = 1 \times 10^{-6} \, m \)

Step 2: Formula for energy density.
Energy density (\( u \)) between the plates of a capacitor is given by:
\[ u = \frac{1}{2} \varepsilon_0 E^2 \] where \( E \) is the electric field and \( \varepsilon_0 = 8.854 \times 10^{-12} \, F/m \).

Step 3: Relation between \( E \), \( V \), and \( d \).
The electric field between the plates is:
\[ E = \frac{V}{d} \] Substitute the given values:
\[ E = \frac{20}{1 \times 10^{-6}} = 2 \times 10^{7} \, V/m. \]

Step 4: Substitute in the energy density formula.
\[ u = \frac{1}{2} \times 8.854 \times 10^{-12} \times (2 \times 10^7)^2 \] \[ u = 0.5 \times 8.854 \times 10^{-12} \times 4 \times 10^{14} \] \[ u = 17.708 \times 10^{2} = 1.77 \times 10^{3} \, J/m^3. \] Approximating gives: \[ u \approx 1.8 \times 10^3 \, J/m^3. \]

Final Answer:
\[ \boxed{1.8 \times 10^3 \, J/m^3} \]