Question

part a
how much energy is stored by the electric field between two square plates, 8.9 cm on a side, separated by a 1.5 - mm air gap? the charges on the plates are equal and opposite and of magnitude 19 nc.
express your answer using two significant figures and include the appropriate units.
pe = value units
submit request answer

Explanation:

Step1: Calculate the area of the plates

The area of a square plate with side length $s = 8.9\ cm=0.089\ m$ is $A = s^{2}$. So $A=(0.089)^{2}= 0.007921\ m^{2}$.

Step2: Calculate the capacitance of the parallel - plate capacitor

The capacitance of a parallel - plate capacitor is given by $C=\frac{\epsilon_{0}A}{d}$, where $\epsilon_{0}=8.85\times 10^{-12}\ F/m$ and $d = 1.5\ mm = 1.5\times10^{-3}\ m$. Then $C=\frac{8.85\times 10^{-12}\times0.007921}{1.5\times 10^{-3}}=4.67\times 10^{-11}\ F$.

Step3: Calculate the energy stored in the capacitor

The energy stored in a capacitor is $U=\frac{Q^{2}}{2C}$, where $Q = 19\ nC=19\times 10^{-9}\ C$. Substitute $Q$ and $C$ into the formula: $U=\frac{(19\times 10^{-9})^{2}}{2\times4.67\times 10^{-11}}=\frac{3.61\times 10^{-16}}{9.34\times 10^{-11}} = 3.9\times10^{-6}\ J$.

Answer:

$3.9\times 10^{-6}\ J$