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Expert-verified Found in: Page 696 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # 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 $$15.0kV$$?

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.

See the step by step solution

## 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}{}\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}{}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 $$3.33\;m$$. ### Want to see more solutions like these? 