Question

A capacitor with a gap of 2 mm has a potential difference from one plate to the other of 30 V. What is the magnitude of the electric field between the plates?

Short answer

The electric field between the two plates is $-15000\hspace{0.33em}\mathrm{N}/\mathrm{C}$.

Step-by-step solution

Electric field between plates

A conductor having two parallel plates with a certain amount of gap consists of an electric field between the plates.

The value of this electric field present between the plates relies on the gap between plates as well as the potential difference between them

Given data

The gap between the capacitor plates is, $\mathrm{\Delta l}=2\hspace{0.33em}\mathrm{mm}\times \frac{1\hspace{0.33em}\mathrm{m}}{1000\mathrm{mm}}=0.002\hspace{0.33em}\mathrm{m}.$

The potential difference between plates is, $\mathrm{\Delta V}=30\hspace{0.33em}\mathrm{V}$.

The electric field between the plates

The formula for the magnitude of the electric field between the two plates of a capacitor is given by,

$$\begin{gathered} \mathrm{E}=\frac{-\mathrm{\Delta V}}{\mathrm{\Delta l}} \\ \mathrm{E}=\frac{-30\hspace{0.33em}\mathrm{V}}{0.002\hspace{0.33em}\mathrm{m}}\times \frac{1\hspace{0.33em}\mathrm{N}/\mathrm{C}}{1\hspace{0.33em}\mathrm{V}/\mathrm{m}} \\ \mathrm{E}=-15000\hspace{0.33em}\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the electric field between the two plates is $-15000\mathrm{N}/\mathrm{C}$.$-15000\mathrm{N}/\mathrm{C}$.