Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

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.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 \(3.33\;m\).

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College Physics (Urone)

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Short Answer

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

Step 1: Principle

Step 2: The given data

Step 3: Calculation of the distance between two plates

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College Physics (Urone)

Answers without the blur.

Short Answer

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

Step 1: Principle

Step 2: The given data

Step 3: Calculation of the distance between two plates

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