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.
Explanation:
The potential/voltage difference between the capacitor plates is given by:
[tex]\begin{gathered} V=Ed \\ \\ E=\frac{V}{d} \\ \\ E=\frac{28.68\text{ V}}{14.05\times10^{-3}\text{ m}} \\ \\ E=2041.28\text{ V/m} \end{gathered}[/tex]Final answer:
The magnitude of the electric field between the plates of the capacitor is 2041.28 V/m.
