Movement through an electric field

Click For Summary
SUMMARY

The discussion focuses on calculating the work done in moving a charge of 11 μC through a constant electric field of 120 V/m from coordinates (-67 m, -36 m) to (98 m, 75 m). Two methods are proposed for solving the problem: first, by calculating the voltage difference and multiplying it by the charge to find the work done; second, by determining the force using the electric field and then applying the work formula (work = force × displacement). Both methods should yield the same result, emphasizing the importance of vector components in the calculations.

PREREQUISITES
  • Understanding of electric fields and their properties
  • Knowledge of voltage and its relationship to work and charge
  • Familiarity with vector mathematics and dot products
  • Basic principles of mechanics, specifically work and force calculations
NEXT STEPS
  • Calculate the voltage difference between two points in an electric field
  • Learn how to apply the formula for work done (work = force × displacement)
  • Explore vector components in electric fields and forces
  • Study the relationship between electric field strength and force on a charge
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in understanding the principles of work done in electric fields.

roanoar
Messages
13
Reaction score
0

Homework Statement



A constant electric field of magnitude E = 120 V/m points in the positive x-direction. How much work (in J) does it take to move the charge Q =11 μC from x1 = -67 m, y1 = -36 m, to x2 = 98 m, y2 = 75 m?

Homework Equations


You see that's the problem... I have no idea


The Attempt at a Solution


Its not much of an attempt but i know work is force times displacement right? So its just the force needed to move it times how far it is moved. But i don't know how to find the force needed to move it
 
Physics news on Phys.org
roanoar said:

Homework Statement



A constant electric field of magnitude E = 120 V/m points in the positive x-direction. How much work (in J) does it take to move the charge Q =11 μC from x1 = -67 m, y1 = -36 m, to x2 = 98 m, y2 = 75 m?

Homework Equations


You see that's the problem... I have no idea


The Attempt at a Solution


Its not much of an attempt but i know work is force times displacement right? So its just the force needed to move it times how far it is moved. But i don't know how to find the force needed to move it

There are two ways to approach this.

First you can determine the voltage difference between the two points. Since voltage is energy (or work) per unit charge, you will take the voltage difference times the charge to find the work done. Remember that voltage is the line integral of electric field over distance (vectorially with dot product).

Second, you can use the fact that electric field is force per unit charge. Hence electric field times charge tells you the force. You already described how to use force to calculate work done.

Try both methods and make sure you get the same answer either way. Remember that force and electric field is a vector, so be careful with direction and vector components.
 

Similar threads

Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
8K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K