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

(a) $d$ = 10 mm and $r$ = 4.24 cm (b) $Q = 5 \times 10^{-10} \mathrm{~ C}$
(a) We want to calculate the dimensions and the separation distance $d$. To determine the dimensions we must find the radius from the area. First, let us find the separated distance where the electric field between the two plates depends on the distance by $$d = \dfrac{V}{E}$$ Where $V$ is the potential. Substitute the values of $V$ and $E$ to find $d$ $$d = \dfrac{1.00 \times 10^{2} \,\text{V} }{1.00 \times 10^{4} \,\text{N/C}} = 10 \,\text{mm}$$ We could use equation 24.2 in the textbook to determine the area $A$ in the form \begin{equation} A = \pi r^2= \dfrac{Cd}{\epsilon_o } \tag{1} \end{equation} Let us substitute the values of $\epsilon_o, C$ and $d$ into equation (1) to get the value of $r$ \begin{align} r &= \sqrt{\dfrac{Cd}{\pi \epsilon_o }}\\ & = \sqrt{\dfrac{(5\times 10^{-12} \,\text{F})(0.01 \,\text{m})}{\pi (8.85 \times 10^{-12} \,\text{F/m})}}\\ &= 4.24 \,\text{cm} \end{align} (b) The capacitance between the two plates occur due to the accumulation of the charges on the plates and it is given by $$Q = CV_{ab}$$ Substitute the values $C$ and $V_{ab}$ to get the charge on the plate $$Q = CV_{ab} = (5 \times 10^{-12} \,\text{F})(1.00 \times 10^{2}\,\text{V}) = 5 \times 10^{-10} \mathrm{~ C}$$