Question

A parallel plate capacitor has a charge of \(0.020 \mu \mathrm{C}\) on each plate with a potential difference of \(240 \mathrm{V}\). The parallel plates are separated by 0.40 mm of air. What energy is stored in this capacitor?

Short answer

Answer: The energy stored in the capacitor is \(2.40 \times 10^{-7} \mathrm{J}\).

Step-by-step solution

Calculate the capacitance

Now we will calculate the capacitance of the capacitor: \(C = \frac{0.020 \times 10^{-6} \mathrm{C}}{240 \mathrm{V}} = 8.33 \times 10^{-11} \mathrm{F}\) The capacitance of the capacitor is \(8.33 \times 10^{-11} \mathrm{F}\). #Step 2: Find the energy stored in the capacitor# Now that we have the capacitance, we can find the energy stored using the formula \(U = \frac{1}{2}CV^2\). We have: \(C = 8.33 \times 10^{-11} \mathrm{F}\) \(V = 240 \mathrm{V}\) Plug in the values, \(U = \frac{1}{2}(8.33 \times 10^{-11} \mathrm{F})(240 \mathrm{V})^2\)

Calculate the energy stored

Now we will calculate the energy stored in the capacitor: \(U = \frac{1}{2}(8.33 \times 10^{-11} \mathrm{F})(240 \mathrm{V})^2 = 2.40 \times 10^{-7} \mathrm{J}\) The energy stored in the capacitor is \(2.40 \times 10^{-7} \mathrm{J}\).