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

(a) $W = 2.2 \times 10^{-6} \,\text{J}$ (b) $V_{a} - V_{b} = 523 \,\text{V}$ (c) $E = 8.71 \times 10^{3} \,\text{V/m}$
(a) The work done is equal to the change in kinetic energies where $K_a = 0$ \begin{align*} W &= \Delta K \\ & = K_b -K_a \\ &=K_b\\ &= \boxed{ 2.2 \times 10^{-6} \,\text{J}} \end{align*} (b) The potential of the starting point with respect to the end point is $$V_a - V_b$$ And it is calcauted by \begin{align*} V_{a} - V_{b} &= \dfrac{W }{q} \\ &= \dfrac{2.20 \times 10^{-6} \,\text{J}}{4.20 \times 10^{-9} \,\text{C}}\\ &= \boxed{523 \,\text{V}} \end{align*} (c) The electric field is related to the difference in potential between the two points \begin{gather*} V_{a} - V_{b} = EL \\ E = \dfrac{ V_{a} - V_{b}}{L} \\ E = \dfrac{523 \,\text{V}}{0.06\,\text{m}}\\ \boxed{E = 8.71 \times 10^{3} \,\text{V/m}} \end{gather*}