A parallel plate electric field has its plates separated by $5\mathrm{.}33$ mm$.$ If $600$ V of potential difference exists between the plates, calculate: $($a$)$ the magnitude of the E$-$Field between the plates.

Correct answer:

$1\mathrm{.}13$E$5$ N$/$C

$($b$)$ the magnitude of the force on an e$-$ between the plates.

Correct answer:

$1\mathrm{.}80$E$-14$ N

$($c$)$ how much work must be done on the e$-$ to move it to the negative plate if it is initially positioned $2\mathrm{.}90$ mm from the positive plate.