Answer
(a) $F = 1.0\times 10^{-6}~N$
(b) $W = 2.5\times 10^{-7}~J$
(c) $\Delta V = 60~J/C$
Work Step by Step
(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:
$\Delta V = E~d$
$\Delta V = (240~N/C)(0.25~m)$
$\Delta V = 60~J/C$
