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

(a) $V_B$ = 80 V (b) $|E| = 100 \,\text{V/m}$ and its direction from $B$ to $A$
(a) For two instants at $A$ and $B$, we apply the conservation law \begin{gather*} K_{A} + U_{A} = K_{B} + U_{B} \\ 0 + qV_{A} = K_{B} + qV_{B} \\ V_{A} = \dfrac{K_{B}}{q} + V_{B} \\ V_{B} = V_{A} - \dfrac{K_{B}}{q} \\ V_B = + 30.0 \,\text{V} - \dfrac{(3.00 \times 10^{-7} \,\text{J})}{(- 6.00 \times 10^{-9} \,\text{C})}\\ \boxed{V_B = + 80.0 \,\text{V}} \end{gather*} (b) The electric field is related to the difference in potential between the two points \begin{gather*} \Delta V = E L\\ V_{A} - V_{B} = EL E = \dfrac{ V_{A} - V_{B}}{L} \\ E = \dfrac{30.0 \,\text{V} - 80.0 \,\text{V}}{0.500 \,\text{m}}\\ E = -100 \,\text{V/m}\\ \boxed{|E| = 100 \,\text{V/m}} \end{gather*} The direction of the electric field is from $B$ to $A$