Question

What is the electric field between the plates of a capacitor that has a charge of 3.97 microC and voltage difference between the plates of 28.68 Volts if the plates are separated by 14.05 mm?

Answer 1

Given:

The charge on the plate of the capacitor is: q = 3.97 microC.

The voltage difference between the plates is: V = 28.68 Volts.

The plates are separated by the distance: d = 14.05 mm.

To find:

The electric field between the plates of a capacitor.

Step-by-step explanation:

The potential/voltage difference between the capacitor plates is given by:

`\begin{gathered} V=Ed \\ \\ E=(V)/(d) \\ \\ E=\frac{28.68\text{ V}}{14.05*10^(-3)\text{ m}} \\ \\ E=2041.28\text{ V/m} \end{gathered}`

Final answer:

The magnitude of the electric field between the plates of the capacitor is 2041.28 V/m.