Question

24.7 • a parallel - plate air capacitor is to store charge of magnitude 240.0 pc on each plate when the potential difference between the plates is 42.0 v. (a) if the area of each plate is 6.80 cm², what is the separation between the plates? (b) if the separation between the two plates is double the value calculated in part (a), what potential difference is required for the capacitor to store charge of magnitude 240.0 pc on each plate?

Explanation:

Step1: Calculate the capacitance using the definition

The definition of capacitance is $C=\frac{Q}{V}$. Given $Q = 240.0\ pC=240\times10^{- 12}\ C$ and $V = 42.0\ V$, we have $C=\frac{240\times10^{-12}}{42.0}\ F\approx5.71\times10^{-12}\ F$.

Step2: Use the formula for the capacitance of a parallel - plate capacitor to find the plate separation

The formula for the capacitance of a parallel - plate capacitor is $C=\frac{\epsilon_0A}{d}$, where $\epsilon_0 = 8.85\times10^{-12}\ F/m$, $A = 6.80\ cm^{2}=6.80\times10^{-4}\ m^{2}$. Rearranging for $d$, we get $d=\frac{\epsilon_0A}{C}$. Substituting the values, $d=\frac{8.85\times10^{-12}\times6.80\times10^{-4}}{5.71\times10^{-12}}\ m\approx1.05\times10^{-3}\ m = 1.05\ mm$.

Step3: Analyze the new situation when the separation is doubled

When the separation $d$ is doubled, the new capacitance $C'=\frac{\epsilon_0A}{2d}=\frac{C}{2}$ (since $C=\frac{\epsilon_0A}{d}$). The charge $Q$ remains the same ($Q = 240\times10^{-12}\ C$). Using $C'=\frac{Q}{V'}$, and since $C'=\frac{C}{2}$, we have $V'=\frac{Q}{C'}=\frac{Q}{\frac{C}{2}} = 2\frac{Q}{C}$. Since $\frac{Q}{C}=V = 42.0\ V$, then $V'=84.0\ V$.

Answer:

(a) $1.05\ mm$
(b) $84.0\ V$