(i) Calculate the potential difference between the plates of a parallel-plate air capacitor. Each plate has a charge of magnitude 0.200 μC, and the capacitance of the capacitor is 300 pF. The plates are separated by a distance of 0.400 mm.
(ii) Determine the area of each plate in the same parallel-plate capacitor described above.
(iii) Compute the magnitude of the electric field between the plates of the capacitor.
(iv) Find the surface charge density on each plate of the capacitor.
Use: $\epsilon_{0}=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}$

• A. $\begin{array}{l}\text{(i) Potential Difference } V=667 \mathrm{~V}\\ \text{ii) Area } A=0.0136 \mathrm{~m}^{2}\\ \text{iii) Electric Field } E=1.67 \times 10^{6} \mathrm{~V} / \mathrm{m}\\ \text{iv) Surface Charge Density } \sigma=14.8 \mu \mathrm{C} / \mathrm{m}^{2}\end{array}$
• B. $\begin{array}{l}\text{(i) Potential Difference }V=667\mathrm{~V}\\\text{ii) Area }A=0.0136~m^2\\\text{iii) Electric Field }E=1.65\times 10^6\mathrm{~V}/\mathrm{m}\\\text{iv) Surface Charge Density }\sigma =14.7\mu \mathrm{C}/\mathrm{m}^2\end{array}$
• C. $\begin{array}{l}\text{i) Potential Difference }V=6567~V\\\text{ii) Area }A=0.0136\mathrm{~m}^2\\\text{iii) Electric Field }E=1.63\times 10^6\mathrm{~V}/\mathrm{m}\\\text{iv) Surface Charge Density }\sigma =15.0\mu \mathrm{C}/\mathrm{m}^2\end{array}$
• D. $\begin{array}{l}\text{i) Potential Difference }V=667\mathrm{~V}\\\text{ii) Area }A=0.0136~m^2\\\text{iii) Electric Field }E=1.70\times 10^6\mathrm{~V}/\mathrm{m}\\\text{iiv) Surface Charge Density }\sigma =14.9\mu \mathrm{C}/\mathrm{m}^2\end{array}$

Answer: (A)