An air-filled 3.0 nF capacitor is charged to 8.0 V.If the plate separation is 100 $\mu$m,what is the energy density in the electric field?

Question

Quizplus · Accepted Answer

The energy density $ u $ in the electric field of a capacitor is given by $ u = \frac{1}{2} \epsilon_0 E^2 $, where $ E = \frac{V}{d} $. For an air-filled capacitor, $ \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} $. Substituting $ V = 8.0 \, \text{V} $ and $ d = 100 \times 10^{-6} \, \text{m} $, we find $ E = 8.0 \times 10^4 \, \text{V/m} $. Thus, $ u = \frac{1}{2} \times 8.85 \times 10^{-12} \times (8.0 \times 10^4)^2 \approx 28 \, \text{mJ/m}^3 $.

An Air-Filled 3 μ\muμ M,what Is the Energy Density in the Electric Field? A)

An Air-Filled 3 $\mu$ M,what Is the Energy Density in the Electric Field?
A)