Answer
$3.9\times10^4V/m$
Work Step by Step
Equation 17–4a tells us that the voltage difference V between the plates is the electric field strength, multiplied by the separation distance, so V = Ed. Use that with equation 17–7.
$$Q=CV=CEd$$
$$E=\frac{Q}{Cd}=\frac{62\times10^{-6}C}{(0.80\times10^{-6}F)(0.0020m)}$$
$$=3.9\times10^4V/m$$
