Question

A capacitor with a gap of $\mathbf{1}\mathbf{mm}$ has a potential difference from one plate to the other of$\mathbf{36}\mathbf{V}$ . What is the magnitude of the electric field between the plates?

Short answer

The magnitude of the electric field between the plates is$36\times {10}^3\text{\hspace{0.17em}V/m}$ .

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The gap between a capacitor plates is, $\mathrm{d}=\left[1\text{\hspace{0.17em}mm\hspace{0.17em}}\times \left(\frac{{10}^{-3}\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}mm}}\right)\right]={10}^{-3}\text{\hspace{0.17em}m}$.
• The potential difference from one plate to the other is, $\mathrm{V}=36\text{\hspace{0.17em}V}$.

Significance of parallel plate capacitor

Whenever two parallel conducting plates lie parallel at a specific distance and connect with a particular potential difference, this arrangement refers to a parallel plate capacitor.

Determination the magnitude of the electric field between the plates

The relation of the magnitude of the electric field between the plates is expressed as,

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

Here, $\mathrm{E}$ is the magnitude of the electric field between the plates.

Substitute all the known values in the above equation.

$\begin{array}{c}\mathrm{E}=\left[\frac{36\mathrm{V}}{{10}^{-3}\text{\hspace{0.17em}m}}\right]\\ =36\times {10}^3\text{\hspace{0.17em}V/m}\end{array}$

Thus, the magnitude of the electric field between the plates is $36\times {10}^3\text{\hspace{0.17em}V/m}$.