Question

A parallel plate capacitor of capacitance \(1\ \mu\text{F}\) is charged to a potential difference of \(20\ \text{V}\). The distance between plates is \(1\ \mu\text{m}\). The energy density between plates of capacitor is:

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

Hint:

For capacitor energy density, first calculate \(E=\frac{V}{d}\), then use \(u=\frac{1}{2}\varepsilon_0E^2\).

Answer 1

The Correct Option is A

Concept:
Energy density in an electric field is: \[ u=\frac{1}{2}\varepsilon_0E^2 \] Also, electric field between plates is: \[ E=\frac{V}{d} \]

Step 1: Write the given values.
Potential difference: \[ V=20\ \text{V} \] Distance between plates: \[ d=1\ \mu\text{m} \] \[ d=1\times 10^{-6}\ \text{m} \]

Step 2: Find electric field.
\[ E=\frac{V}{d} \] \[ E=\frac{20}{1\times 10^{-6}} \] \[ E=2\times 10^7\ \text{V/m} \]

Step 3: Find energy density.
\[ u=\frac{1}{2}\varepsilon_0E^2 \] Using: \[ \varepsilon_0=8.85\times 10^{-12}\ \text{F/m} \] \[ u=\frac{1}{2}(8.85\times 10^{-12})(2\times 10^7)^2 \] \[ u=\frac{1}{2}(8.85\times 10^{-12})(4\times 10^{14}) \] \[ u=\frac{1}{2}(35.4\times 10^2) \] \[ u=17.7\times 10^2 \] \[ u=1.77\times 10^3\ \text{J/m}^3 \] \[ u\approx 1.8\times 10^3\ \text{J/m}^3 \]

Step 4: Final conclusion.
Hence, the energy density is: \[ \boxed{1.8\times 10^3\ \text{J/m}^3} \]