Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

The naturally occurring charge on the ground on a fine day out in the open country is \( - 1.00n{\rm{ }}C/{m^2}\).
(a) What is the electric field relative to ground at a height of \(3.00{\rm{ }}m\)?
(b) Calculate the electric potential at this height.
(c) Sketch electric field and equipotential lines for this scenario.

(a) The electric field relative to ground is \(E = - 112.94{\rm{ }}V/m\).

(b) The electric potential is \(V = - 338.83\;V\).

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Information

Charge on the ground \( - 1.00n{\rm{ }}C/{m^2}\)
Height at which electric field is – \(3.00{\rm{ }}m\)

Step 2: Concept Introduction

An electric field is an electric characteristic associated with any location in space where a charge, in any form, is present. An alternative name for an electric field is the electric force per unit charge.

Equipotential Lines: Equipotential lines are traces of equal-altitude lines, like contour lines on a map. In this instance, electric potential or voltage serves as the "height." Equipotential lines are always perpendicular to the electric field.

Step 3: Electric Field

(a)

Calculate the electric field and potential of flat ground with charge density \(\sigma = - 1n{\rm{ }}C/{m^2}\).

Consider the ground as a uniformly charged infinite plate. The electric field in this case is constant and points towards the ground. The field is equal to –

\(\begin{array}{}E &= \frac{\sigma }{{{ \in _0}}}\\ &= \frac{{ - {{10}^{ - 9}}{\rm{ }}C/{m^2}}}{{8.854 \times {{10}^{ - 12}}{\rm{ }}C/Vm}}\\ &= - 112.94\;V/m\end{array}\)

Step 4: Electric Potential

(b)

The potential at \(d = 3{\rm{ }}m\) distance is obtained as –

\(\begin{array}{}V &= Ed\\ &= (112.94\;V/m) \times (3\;m)\\ &= - 338.83\;V\end{array}\)

Step 5: Plotting Electric Field and Equipotential Lines

(c)

The field lines are perpendicular to the ground and point towards the ground. Since they must be perpendicular to the field lines in order to qualify as equipotential lines, they are surfaces parallel to the ground. The following figure shows this.

Hence, the electric field and electric potential is \(E = - 112.94{\rm{ }}V/m\) and \(V = - 338.83\;V\)respectively. The electric fields and equipotential lines are also plotted.

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Given Information

Step 2: Concept Introduction

Step 3: Electric Field

Step 4: Electric Potential

Step 5: Plotting Electric Field and Equipotential Lines

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College Physics (Urone)

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Given Information

Step 2: Concept Introduction

Step 3: Electric Field

Step 4: Electric Potential

Step 5: Plotting Electric Field and Equipotential Lines

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