Question

What is the electric field between the plates of a capacitor that has a charge of 14.85 microC and voltage difference between the plates of 59.98 Volts if the plates are separated by 2.17 mm?

Answer 1

The electric field between two parallel plates is given by the equation:

$E=\frac{V}{d}$

where V is the difference of potential and d is the distance between the plates. In this case the difference of potential is 59.98 V and the distance between the plates is 2.17 mm, then we have:

$\begin{gathered} E=\frac{59.98}{2.17\times10^{-3}} \\ E=2.76\times10^4 \end{gathered}$

Therefore, the electric field is:

$E=2.76\times10^4\text{ }\frac{N}{C}$