#### Answer

(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.

#### Work Step by Step

(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.