## University Physics with Modern Physics (14th Edition)

(a) $Q = 5 \times 10^{-9} \mathrm{~ C}$ (b) $E = 1.2 \times 10^{4} \,\text{N/C}$ (c) $E_i = 2.5 \times 10^{4} \,\text{N/C}$
(a) First, let us find the capacitance by $$C = \dfrac{K \epsilon_o A}{d} = \dfrac{(2.1) (8.85 \times 10^{-12} \,\text{F/m}) (0.0225 \mathrm{~m^2})}{ 1 \times 10^{-3} \,\text{m}} = 418 \times 10^{-12} \,\text{F}$$ After calculating the capacitance we could determine the charges where the capacitance between the two plates occur due to the accumulation of the charges on the plates because of the potential of the battery and it is given by $$Q = CV_{ab}$$ Where $V_{ab}$ is the potential between the two plates. Let us substitute the values $C$ and $V_{ab}$ to get the charge on each plate $$Q = CV_{ab} = (418\times 10^{-12} \,\text{F})(12 \,\text{V}) = \boxed{5 \times 10^{-9} \mathrm{~ C}}$$ (b) We integrate equation 24.23 for $A$ to determine the electric field by \begin{gather*} \oint K \vec{E} \cdot d\vec{A} = \dfrac{Q}{\epsilon_o} \\ K E A = \dfrac{Q}{\epsilon_o} \\ E = \dfrac{Q}{\epsilon_o K A}\\ E = \dfrac{5 \times 10^{-9} \mathrm{~ C}}{(8.85 \times 10^{-12} \,\text{F/m}) (2.1) (0.0225 \mathrm{~m^2})} \\ \boxed{E = 1.2 \times 10^{4} \,\text{N/C}} \end{gather*} (c) The electric field will be calculated with the same steps but without $K$ and could be calculated by \begin{gather*} E_i = \dfrac{Q}{\epsilon_o A}\\ E_i = \dfrac{5 \times 10^{-9} \mathrm{~ C}}{(8.85 \times 10^{-12} \,\text{F/m}) (0.0225 \mathrm{~m^2})} \\ \boxed{E_i = 2.5 \times 10^{4} \,\text{N/C}} \end{gather*}