Suggested languages for you:

Americas

Europe

16PE

Expert-verifiedFound in: Page 696

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.

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

- 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:

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

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 \(3.33\;m\).

94% of StudySmarter users get better grades.

Sign up for free