University Physics with Modern Physics (14th Edition)

(a) $C$ = 15 pF (b) $r_a$ = 3 cm (c) $E = 3.3 \times 10^{4} \,\text{N/C}$
(a) The capacitance between the two plates occur due to the accumulation of the charges on the plates because of the potential difference and it is given by $$C = \dfrac{ Q}{V_{ab}}$$ Where $V_{ab}$ is the potential between the two plates. Let us substitute the values $Q$ and $V_{ab}$ to get the capacitance $$C = \dfrac{ Q}{V_{ab}} = \dfrac{3.30 \times 10^{-9} \,\text{C}}{220 \,\text{V}} = 15 \,\text{pF}$$ (b) The capacitance for a spherical capacitor is calculated by \begin{gather*} C = 4 \pi \epsilon_o \dfrac{r_a r_b}{r_b - r_a}\\ C = 4 \pi \epsilon_o \dfrac{1 }{\dfrac{1}{r_a} - \dfrac{1}{r_b}}\\ \dfrac{1}{r_a} - \dfrac{1}{r_b} = \dfrac{4 \pi \epsilon_o }{C} \\ \dfrac{1}{r_a} = \dfrac{4 \pi \epsilon_o }{C} + \dfrac{1}{r_b} \end{gather*} Substitute for $C, \epsilon_o$ and $r_b$ \begin{align*} \dfrac{1}{r_a} &= \dfrac{4 \pi \epsilon_o }{C} + \dfrac{1}{r_b} \\ &= \dfrac{4 \pi (8.85 \times 10^{-12} \,\text{F/m}) }{15 \times 10^{-12} \,\text{F}} + \dfrac{1}{0.04 \,\text{m}}\\ &= 32.4 \mathrm{~m^{-1}} \end{align*} Now take the reciprocal of the answer $$r_a = \dfrac{1}{32.4 \mathrm{~m^{-1}}} = 3 \mathrm{~cm}$$ (c) The electric field of the sphere is given by Gauss's law in the form \begin{align*} E = \dfrac{1}{4\pi \epsilon_o} \dfrac{Q}{r_a^2} \end{align*} Substitute to get $E$ \begin{align*} E = \dfrac{1}{4\pi \epsilon_o} \dfrac{3.30 \times 10^{-9} \,\text{C}}{( 0.03 \mathrm{~m})^2}= 3.3 \times 10^{4} \,\text{N/C} \end{align*}