## College Physics (4th Edition)

(a) $F = 1.0\times 10^{-6}~N$ (b) $W = 2.5\times 10^{-7}~J$ (c) $\Delta V = 60~J/C$
(a) We can find the force exerted on the particle: $F = E~q$ $F = (240~N/C)(4.2\times 10^{-9}~C)$ $F = 1.0\times 10^{-6}~N$ (b) We can find the work done by the electric field: $W = F~d$ $W = (1.0\times 10^{-6}~N)(0.25~m)$ $W = 2.5\times 10^{-7}~J$ (c) We can find the potential difference: $\Delta V = E~d$ $\Delta V = (240~N/C)(0.25~m)$ $\Delta V = 60~J/C$