Question

a parallel - plate capacitor has a plate area of 0.30 m² and a plate separation of 0.10 mm. if the charge on each plate has a magnitude of 5.0 x 10⁻⁶ c, what is the energy density in its electric field? o 24 j/m³ o 0.16 j/m³ o 7.8 j/m³ o 3.5 j/m³ o 16 j/m³ save for later submit answer 25.4 energy stored in an electric field

Explanation:

Step1: Calculate the capacitance

The capacitance of a parallel - plate capacitor is given by $C=\frac{\epsilon_0A}{d}$, where $\epsilon_0 = 8.85\times10^{-12}\ F/m$, $A = 0.30\ m^2$ and $d=0.10\times10^{-3}\ m$.
$C=\frac{8.85\times10^{-12}\times0.30}{0.10\times10^{-3}}=2.655\times10^{-7}\ F$

Step2: Calculate the potential difference

We know that $Q = 5.0\times10^{-6}\ C$ and $Q = CV$, so $V=\frac{Q}{C}$.
$V=\frac{5.0\times10^{-6}}{2.655\times10^{-7}}\approx18.83\ V$

Step3: Calculate the electric - field strength

The electric - field strength in a parallel - plate capacitor is $E=\frac{V}{d}$.
$E=\frac{18.83}{0.10\times10^{-3}} = 1.883\times10^{5}\ V/m$

Step4: Calculate the energy density

The energy density in an electric field is $u=\frac{1}{2}\epsilon_0E^{2}$.
$u=\frac{1}{2}\times8.85\times10^{-12}\times(1.883\times10^{5})^{2}$
$u=\frac{1}{2}\times8.85\times10^{-12}\times3.545\times10^{10}$
$u = 0.16\ J/m^{3}$

Answer:

$0.16\ J/m^{3}$