Question

how much energy is stored by the electric field between two square plates, 7.5 cm on a side, separated by a 2.5 - mm air gap? the charges on the plates are equal and opposite and of magnitude 11 nc. express your answer using two significant figures and include the appropriate units.

Explanation:

Step1: Calculate the area of the plates

The side length of the square plate \(s = 7.5\ cm=0.075\ m\). The area of a square \(A = s^{2}\).

$$A=(0.075)^{2}= 5.625\times10^{-3}\ m^{2}$$

Step2: Calculate the capacitance \(C\)

The formula for the capacitance of a parallel - plate capacitor is \(C=\frac{\epsilon_{0}A}{d}\), where \(\epsilon_{0}=8.85\times 10^{-12}\ F/m\) and \(d = 2.5\ mm = 2.5\times10^{-3}\ m\)

$$C=\frac{8.85\times 10^{-12}\times5.625\times 10^{-3}}{2.5\times 10^{-3}}=1.99125\times10^{-11}\ F$$

Step3: Calculate the stored energy \(PE\)

The formula for the energy stored in a capacitor is \(PE=\frac{Q^{2}}{2C}\), where \(Q = 11\ nC=11\times10^{-9}\ C\)

$$PE=\frac{(11\times 10^{-9})^{2}}{2\times1.99125\times 10^{-11}}=\frac{121\times10^{-18}}{3.9825\times10^{-11}}$$

$$PE = 3.04\times10^{-6}\ J\approx3.0\times 10^{-6}\ J$$

Answer:

\(3.0\times 10^{-6}\ J\)