Electric Potential Difference and the Electric Field As you saw in the Electric Potential Lab, you can relate the Electric Field of a system to a map of the potential. Mathematically we can do this as well when we have a function for the Electric Potential Difference in all space. There is a function below for the Electric Potential Difference. The letters A, B and C are constant values shown below the function. AV = Ax²y² + Bxz³ – Cxyz² || - A = -0.20 V/m B = -5.20 V/m4 C = -3.7 V/m4 %3D For this problem, you need to find the component of the Electric Field in the x direction at a position (2.90 m, 2.50 m, -1.30 m). Be careful that you do the math in the correct order (i.e. - don't plug in the position first). Make sure to put units with your answer and include a negative sign if your answer gives you one. Make sure to give your answer with an appropriate number of significant figures. There are 2 correct ways to write these units, but D2L can only store 1 way. Be aware of that when checking your answer-I'll go in and manually change the points. Your Answer: Answer units
Electric Potential Difference and the Electric Field As you saw in the Electric Potential Lab, you can relate the Electric Field of a system to a map of the potential. Mathematically we can do this as well when we have a function for the Electric Potential Difference in all space. There is a function below for the Electric Potential Difference. The letters A, B and C are constant values shown below the function. AV = Ax²y² + Bxz³ – Cxyz² || - A = -0.20 V/m B = -5.20 V/m4 C = -3.7 V/m4 %3D For this problem, you need to find the component of the Electric Field in the x direction at a position (2.90 m, 2.50 m, -1.30 m). Be careful that you do the math in the correct order (i.e. - don't plug in the position first). Make sure to put units with your answer and include a negative sign if your answer gives you one. Make sure to give your answer with an appropriate number of significant figures. There are 2 correct ways to write these units, but D2L can only store 1 way. Be aware of that when checking your answer-I'll go in and manually change the points. Your Answer: Answer units
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:Electric Potential Difference and the Electric Field
As you saw in the Electric Potential Lab, you can relate the Electric Field of a system
to a map of the potential. Mathematically we can do this as well when we have a
function for the Electric Potential Difference in all space. There is a function below
for the Electric Potential Difference. The letters A, B and C are constant values
shown below the function.
AV = Ax²y² + Bxz³ – Cxyz²
-
A = -0.20 V/m4
B = -5.20 V/m4
C = -3.7 V/m4
For this problem, you need to find the component of the Electric Field in the x
direction at a position (2.90 m, 2.50 m, -1.30 m). Be careful that you do the math in
the correct order (i.e. - don't plug in the position first).
Make sure to put units with your answer and include a negative sign if your answer
gives you one. Make sure to give your answer with an appropriate number of
significant figures.
* There are 2 correct ways to write these units, but D2L can only store 1 way. Be
aware of that when checking your answer - I'll go in and manually change the points.
Your Answer:
Answer
units
Expert Solution
Step 1
Given data
The expression for the potential difference is
The constant value is A = - 0.20 V/m4, B = - 5.20 V/m4 , C = - 3.7 V/m4
The position is given as: x = 2.90 m, y = 2.50 m, z = - 1.30 m
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