16PE - College Physics (Urone)

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 \(15.0kV\)?

Step-by-Step Solution

Answer

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

Step 1: Principle

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

Step 2: 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:

      \(\begin{aligned}{\underline{\phantom{xx}}}\Delta V &= (15.0kV)\left( {\frac{{1000\;V}}{{1kV}}} \right)\\ &= 1.50 \times {10^4}\;V\end{aligned}\).

Step 3: Calculation of the distance between two plates

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

\(d = \frac{{\Delta V}}{E}\)

 

Entering the values for \(\Delta V\) and \(E\), we obtain:

\(\begin{aligned}{\underline{\phantom{xx}}}d &= \frac{{1.50 \times {{10}^4}\;V}}{{4.50 \times {{10}^3}\;V/m}}\\ &= 3.33\;m\end{aligned}\)

 

Therefore, the distance between two plates is \(3.33\;m\).