Question

How far apart are two conducting plates that have an electric field strength of $4.50\times {10}^{3}V/m$ between them, if their potential difference is $15.0kV$?

Short answer

The two conducting plates that have an electric field strength of $4.50\times {10}^{3}V/m$ between them is $3.33m$apart.

Step-by-step solution

Principle

The potential difference between two points separated by a distance$d$in a homogeneous electric field of magnitude$E$is$\mathrm{\Delta }V=Ed......(1)$.

The given data

• The electric field strength between the two plates is:

$E=4.50\times {10}^{3}V/m$.

• The potential difference between the two plates is:

$\begin{array}{rl}\mathrm{\Delta }V& =(15.0kV)\left(\frac{1000V}{1kV}\right)\\ & =1.50\times {10}^{4}V\end{array}$.

Calculation of the distance between two plates

The distance between the two plates is found by solving Equation$(1)$for$d$:

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

Entering the values for$\mathrm{\Delta }V$and$E$, we obtain:

$\begin{array}{rl}d& =\frac{1.50\times {10}^{4}V}{4.50\times {10}^{3}V/m}\\ & =3.33m\end{array}$

Therefore, the distance between two plates is $3.33m$.