Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

What is the magnitude and direction of an electric field that exerts a $2.00 \times 10^{- 5} N$ upward force on a $- 1.75 μ C a$ charge?

The magnitude of the electric field is $11.43 N / C$ , and the direction is downward.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Data

Electric field E exerts a force $F (= 2.00 \times 10^{- 5} N)$ on a charge $(= - 1.75 μ C)$ .

Step 2: Electric field

The force per unit positive test charge is known as the intensity of the electric field.

Step 3: Magnitude and direction of electric field

From the definition of the electric field,

$E = \frac{F}{q}$

Here, F is the force $(F = 2.00 \times 10^{- 5} N)$ , and q is the test charge $(q = 1.75 μ C)$ .

Substituting all known values,

$E = \frac{2.00 \times 10^{- 5} N}{- 1.75 μ C} = \frac{2.00 \times 10^{- 5} N}{(- 1.75 μ C) \times (\frac{10^{- 6} C}{1 μ C})} = 11.43 N / C$

Hence, the magnitude of the electric field is 11.43 N/C.

The negative sign of the electric field indicates that the direction of electric field is in the opposite direction of force. Since, the force is directed upward. Hence, the electric field is directed downward.

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College Physics (Urone)

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Short Answer

What is the magnitude and direction of an electric field that exerts a $2.00 \times 10^{- 5} N$ upward force on a $- 1.75 μ C a$ charge?

Step by Step Solution

Step 1: Given Data

Step 2: Electric field

Step 3: Magnitude and direction of electric field

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College Physics (Urone)

Answers without the blur.

Short Answer

What is the magnitude and direction of an electric field that exerts a 2.00×10-5 N upward force on a -1.75 μCa charge?

Step by Step Solution

Step 1: Given Data

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Step 3: Magnitude and direction of electric field

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