Question

A parallel plate capacitor has a charge of \(5.5 \times 10^{-7} \mathrm{C}\) on one plate and \(-5.5 \times 10^{-7} \mathrm{C}\) on the other. The distance between the plates is increased by \(50 \%\) while the charge on each plate stays the same. What happens to the energy stored in the capacitor?

Short answer

When the distance between the plates is increased by 50%, the energy stored in the parallel plate capacitor increases by a factor of 3/2, or 50%. This is because the capacitance decreases as the distance between the plates is increased, resulting in a higher energy storage.

Step-by-step solution

Finding the initial capacitance of the capacitor

To find the initial capacitance, we will use the formula of capacitance for a parallel plate capacitor: \(C = \frac{\epsilon_0 A}{d}\), where \(C\) is the capacitance, \(\epsilon_0\) is the vacuum permittivity (\(8.85 \times 10^{-12} F/m\)), \(A\) is the area of the plates, and \(d\) is the distance between the plates. However, we need only the charge and distance between the plates which varies. Let's denote the initial distance between the plates as \(d_1\) and let \(C_1\) be the initial capacitance. Considering the charge on each plate remains constant, we know, \(Q = C_1 V_1\), where \(Q\) is the charge on each plate, and \(V_1\) is the potential difference across the capacitor.

Finding the capacitance after the distance is increased

As the distance between the plates increased by \(50\%\), the new distance \(d_2 = 1.5 d_1\). Let the new capacitance be \(C_2\). We know that capacitance is inversely proportional to the distance between the plates, so: \(C_2 = \frac{\epsilon_0 A}{d_2}\), Since \(d_2=1.5d_1\), we get \(C_2 = \frac{C_1}{1.5} =\frac{2}{3}C_1\).

Finding the energy stored in the capacitor initially

Initially, the energy stored in the capacitor can be calculated using the formula: \(E_1=\frac{1}{2}C_1 V_1^2\) Since, \(Q = C_1 V_1\), we can rewrite this expression as: \(E_1=\frac{Q^2}{2C_1}\)

Finding the energy stored in the capacitor after the distance is increased

When the distance between the plates is increased, the charge on each plate remains constant. However, the capacitance changes. The energy stored in the capacitor after the distance is increased can be calculated as: \(E_2=\frac{Q^2}{2C_2}\) Now, to find what happens to the energy stored in the capacitor, let's find the ratio of the energy stored initially to energy stored after the distance is increased.

Comparing the initial and final energy stored in the capacitor

To find the ratio of the initial and final energy, we will divide \(E_2\) by \(E_1\): \(\frac{E_2}{E_1} = \frac{\frac{Q^2}{2C_2}}{\frac{Q^2}{2C_1}}= \frac{C_1}{C_2}\) We already derived that \(C_2 =\frac{2}{3}C_1\), so we get: \(\frac{E_2}{E_1} = \frac{C_1}{\frac{2}{3}C_1}=\frac{1}{\frac{2}{3}}= \frac{3}{2}\) As a result, the energy stored in the capacitor increases by a factor of \(\frac{3}{2}\), or \(50\%\), when the distance between the plates is increased by \(50\%\).