## College Physics (7th Edition)

(a) If the distance from the charge is doubled the magnitude of electric field (2) decreased to the initial value (b) magnitude of new electric field at twice the distance will be $E_{2r}=2.5\times10^{-15}N/C$
(a) From the definition of the electric field magnitude of the electric field due to a point charge is $E= \frac{kq}{r^2}$, so electric field is inversely proportional to square of distance from the charge. So if distance increases electric field decreases. (b) given that magnitude of electric field at distance $r$ is $E_{r}= \frac{kq}{r^2}= 1.0\times10^{-14}N/C$ now if the distance is doubled electric field will be $E_{2r}= \frac{kq}{(2r)^2}=\frac{1}{4}\times \frac{kq}{r^2}$ $E_{2r}=\frac{1}{4}\times1.0\times10^{-14}N/C=0.25\times10^{-14}N/C$ $E_{2r}=2.5\times10^{-15}N/C$