## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) The electric field has a strength of $1.4\times 10^{-3}~N/C$ directed away from the proton. (b) The electric field has a strength of $1.4\times 10^{-3}~N/C$ directed toward the electron.
(a) We can find the magnitude of the electric field 1.0 mm from the proton: $E = \frac{k~q}{r^2}$ $E = \frac{(9.0\times 10^9~N~m^2/C^2)(1.6\times 10^{-19}~C)}{(0.0010~m)^2}$ $E = 1.4\times 10^{-3}~N/C$ An electric field points away from a positive charge. Therefore, the electric field has a strength of $1.4\times 10^{-3}~N/C$ directed away from the proton. (b) We can find the magnitude of the electric field 1.0 mm from the electron: $E = \frac{k~q}{r^2}$ $E = \frac{(9.0\times 10^9~N~m^2/C^2)(1.6\times 10^{-19}~C)}{(0.0010~m)^2}$ $E = 1.4\times 10^{-3}~N/C$ An electric field points toward a negative charge. Therefore, the electric field has a strength of $1.4\times 10^{-3}~N/C$ directed toward the electron.