Part A Determine the magnitude of the electric field E at the origin 0 in the figure(Figure 1) due to the two charges at A and B. Express your answer in terms of the variables Q, I, k, and appropriate constants. ΑΣφ ? E1 = Submit Request Answer • Part B Determine the direction of the electric field E at the origin 0 in the figure due to the two charges at A and B.

University Physics Volume 2
18th Edition
ISBN:9781938168161
Author:OpenStax
Publisher:OpenStax
Chapter5: Electric Charges And Fields
Section: Chapter Questions
Problem 101P: In this exercise, you practice electric field lines. Make sure you represent both the magnitude and...
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Part A
Determine the magnitude of the electric field E at the origin 0 in the figure(Figure 1) due to the two charges at A and B.
Express your answer in terms of the variables Q, l, k, and appropriate constants.
?
E1 =
Submit
Request Answer
Part B
Determine the direction of the electric field E at the origin 0 in the figure due to the two charges at A and B.
?
01 =
counterclockwise from +x axis
Figure
1 of 1
Submit
Request Answer
y
Part C
+Q
A
Repeat A, but let the charge at B be reversed in sign.
Express your answer in terms of the variables Q, 1, k, and appropriate constants.
+Q
να ΑΣΦ
B
?
E2 =
圓
Transcribed Image Text:Part A Determine the magnitude of the electric field E at the origin 0 in the figure(Figure 1) due to the two charges at A and B. Express your answer in terms of the variables Q, l, k, and appropriate constants. ? E1 = Submit Request Answer Part B Determine the direction of the electric field E at the origin 0 in the figure due to the two charges at A and B. ? 01 = counterclockwise from +x axis Figure 1 of 1 Submit Request Answer y Part C +Q A Repeat A, but let the charge at B be reversed in sign. Express your answer in terms of the variables Q, 1, k, and appropriate constants. +Q να ΑΣΦ B ? E2 = 圓
Expert Solution
Step 1

Part (a)

The diagram is as shown below:

Physics homework question answer, step 1, image 1

The expression for electric field in x direction is:

Ex=EBOsin60°Ex=-kQl2sin60°Ex=-32kQl2

The expression for electric field in y direction is:

Ey=-EAO+EBOcos60°Ey=-kQl2+kQl2cos60°Ey=-32kQl2

 

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OpenStax