Question

25.2 calculating the capacitance a parallel - plate capacitor has a plate area of 0.2 m² and a plate separation of 0.1 mm. if the charge on each plate has a magnitude of 4 x 10⁻⁶ c the potential difference across the plates is approximately o 2 x 10² v o 0 v o 4 x 10⁻² v o 2 x 10⁵ v o 4 x 10⁸ v save for later submit answer

Explanation:

Step1: Calculate the capacitance formula

The capacitance of a parallel - plate capacitor is given by $C=\frac{\epsilon_0A}{d}$, where $\epsilon_0 = 8.85\times10^{-12}\ F/m$, $A = 0.2\ m^2$, and $d=0.1\ mm=0.1\times10^{-3}\ m$.
$C=\frac{8.85\times10^{-12}\times0.2}{0.1\times10^{-3}}=1.77\times10^{-8}\ F$

Step2: Use the relationship between charge, capacitance and potential difference

The relationship between charge $Q$, capacitance $C$ and potential difference $V$ is $Q = CV$. We know $Q = 4\times10^{-6}\ C$ and $C=1.77\times10^{-8}\ F$. Then $V=\frac{Q}{C}$.
$V=\frac{4\times10^{-6}}{1.77\times10^{-8}}\approx2\times10^{2}\ V$

Answer:

$2\times10^{2}\ V$