Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

Answers without the blur.

Just sign up for free and you're in.

Short Answer

What is the magnitude and direction of the force exerted on a $3.50 μ C$ charge by a 250 N/C electric field that points due east?

The magnitude of the force is $8.75 \times 10^{- 4} N$ and it is directed towards East.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Electric field

A space around the charge in which any other test charge experiences some force is known as electric field. The expression for the strength of the electric field is,

$E = \frac{F}{q} - - - - - - - - - - (1.1)$

Here, F is the force experienced by the test charge, q and is the test charge.

Step 2: Magnitude and direction of force

The magnitude of the force experienced by the test charge can be calculated using equation (1.1).

Rearranging equation (1.1) in order to get an expression for the force experienced by the test charge.

F = qE

Here, q is the charge on test charge $(q = 3.50 μ C)$ , and E is the strength of electric field (250 N/C).

Substituting all known values,

$F = (3.50 μ C) \times (250 N / C) = (3.50 μ C) \times (\frac{10^{- 6} C}{1 μ C}) \times (250 N / C) = 8.75 \times 10^{- 4} N$

Hence, the magnitude of the force is $8.75 \times 10^{- 4} N$ .

Since, the direction of the force and the electric field are same. Hence, the force is directed towards East.

Find Study Materials for

Create Study Materials

Select your language

College Physics (Urone)

Answers without the blur.

Short Answer

What is the magnitude and direction of the force exerted on a $3.50 μ C$ charge by a 250 N/C electric field that points due east?

Step by Step Solution

Step 1: Electric field

Step 2: Magnitude and direction of force

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

College Physics (Urone)

Answers without the blur.

Short Answer

What is the magnitude and direction of the force exerted on a 3.50 μC charge by a 250 N/C electric field that points due east?

Step by Step Solution

Step 1: Electric field

Step 2: Magnitude and direction of force

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics

What is the magnitude and direction of the force exerted on a $3.50 μ C$ charge by a 250 N/C electric field that points due east?