Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

If the potential due to a point charge is $5.00 \times 10^{2} V$ at a distance of 15.0 m, what are the sign and magnitude of the charge?

The potential V is positive, and the sign of the charge Q is also positive which is obtained as: $Q = 8.33 \times 10^{- 3} C$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Data

Potential due to the charge is $5.00 \times 10^{2} V$ .

Distance from the charge is 15.0 m.

Step 2: Define Electric Field

A physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field, either attractive or repulsive, is referred to as an electric field.

Step 3: The given values

Finding the magnitude and sign of the charge Q, if the potential is due to a point charge is: $V = 5.00 x 10 V$ ,

It is at a distance of r which is: r = 15.0 m.

Step 4: Obtaining the value of :

In the equation of the potential V is: $V = \frac{kQ}{r}$

Determine the charge Q, as: $Q = \frac{rV}{k}$

The value of k= $k = 9.00 \times 10^{9} N . \frac{m^{2}}{C^{2}}$ .

Step 5: Evaluation of the charge

Substituting the quantities in the equation for the charge Q as:

$\begin{array}{rcl} Q & = & \frac{rV}{k} \\ = & \frac{(15.0 m) (5.00 \times 10^{2} V)}{9.00 \times 10^{9} N . \frac{m^{2}}{C^{2}}} \\ = & 8.33 \times 10^{- 3} C \end{array}$

Therefore, the potential V is positive and the sign of the charge Q is positive which gives us: $Q = 8.33 \times 10^{- 3} C$ .

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College Physics (Urone)

Answers without the blur.

Short Answer

If the potential due to a point charge is $5.00 \times 10^{2} V$ at a distance of 15.0 m, what are the sign and magnitude of the charge?

Step by Step Solution

Step 1: Given Data

Step 2: Define Electric Field

Step 3: The given values

Step 4: Obtaining the value of :

Step 5: Evaluation of the charge

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College Physics (Urone)

Answers without the blur.

Short Answer

If the potential due to a point charge is 5.00×102 V at a distance of 15.0 m, what are the sign and magnitude of the charge?

Step by Step Solution

Step 1: Given Data

Step 2: Define Electric Field

Step 3: The given values

Step 4: Obtaining the value of :

Step 5: Evaluation of the charge

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