## College Physics (4th Edition)

(a) The point charge is positive. (b) The electric potential is $+2.0~kV$ at a distance of $10.0~cm$
(a) The electric potential can be expressed as $V = \frac{k~q}{r}$, where $q$ is the point charge. Since the electric potential $V$ is positive, then the point charge $q$ must also be positive. (b) We can write an expression for $V_1$: $V_1 = \frac{k~q}{r_1} = 1.0~kV$ We can write an expression for $V_2$: $V_2 = \frac{k~q}{r_2} = 2.0~kV$ We can find $r_2$ by dividing $V_2$ by $V_1$: $\frac{V_2}{V_1} = \frac{\frac{k~q}{r_2}}{\frac{k~q}{r_1}}$ $\frac{V_2}{V_1} = \frac{r_1}{r_2}$ $r_2 = \frac{V_1~r_1}{V_2}$ $r_2 = \frac{(1.0~kV)(20.0~cm)}{2.0~kV}$ $r_2 = 10.0~cm$ The electric potential is $+2.0~kV$ at a distance of $10.0~cm$