# Magnitude and Direction of Electric Fields

by cse63146
Tags: direction, electric, fields, magnitude
 P: 452 1. The problem statement, all variables and given/known data A small object A, electrically charged, creates an electric field. At a point P located 0.250m directly north of A, the field has a value of 40.0N/C directed to the south. A) 1.11×10^−9C B) −1.11×10^−9C C) 2.78×10^−10C D) −2.78×10^−10C E) 5.75×10^12C F) −5.75×10^12C What is the charge of object A? 2. Relevant equations E = (KQ)/r^2 3. The attempt at a solution After manipulating the above equation, I found out that the charge Q = (E*r^2)/K Q = (E*r^2)/K = (40*(0.25^2))/9×10^9 = 2.78×10^−10C makes sense to me, but should it be negative or positive?
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P: 1,664
 Quote by cse63146 Q = (E*r^2)/K = (40*(0.25^2))/9*10^9 = 2.78*10^-10C makes sense to me, but should it be negative or positive?
Remember that the vector direction of electric field is defined as radially outward from a positive charge and radially inward from a negative charge. The point P is north of the charge and the field vector there points back southward toward the charge. So what sign would the charge need to have? (BTW, I agree with your charge magnitude.)
 P: 662 A way to remember the rule that dynamicsolo stated is to recall that E fields are defined in terms of the force felt by a unit test charge, which is positive by assumption. That gives the directions stated above.
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P: 1,664

## Magnitude and Direction of Electric Fields

 Quote by belliott4488 A way to remember the rule that dynamicsolo stated is to recall that E fields are defined in terms of the force felt by a unit test charge, which is positive by assumption. That gives the directions stated above.
Thanks, I did omit to say that, since it is properly part of that definition.
 P: 452 so it would be positive?
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Thanks
P: 25,175
 Quote by cse63146 so it would be positive?
Why would you think that?
 P: 662 cse63146 - Think it through in these steps: - Is the test charge (i.e. field point in question) North or South of the field source at A? - Given that location, does the direction of the force it feels indicate an attractive or repulsive force? - Considering that the test charge is defined to be positive, what does your previous answer tell you about the sign of the source charge at A?
 P: 452 Since there's a point P north of charge A, and the electric field is directed to the south (towards the charge), it looks like it's inward, which makes the charge negative?
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P: 1,664
 Quote by cse63146 Since there's a point P north of charge A, and the electric field is directed to the south (towards the charge), it looks like it's inward, which makes the charge negative?
That would be correct.
 P: 452 There's a part B of the question which only showed up after you answer part A. If a second object B with the same charge as A is placed at 0.250m south of A (so that objects A and B and point P follow a straight line), what is the magnitude of the total electric field produced by the two objects at P? I know that E = (KQ)/r^2 so E = (KQ)/r^2 E = ((9×10^9)(2.78×10^−10)/0.5^2 E = 10 So the total magnitude produced by the two objects 10 + 40 = 50N/C?
P: 3
 Quote by cse63146 There's a part B of the question which only showed up after you answer part A. If a second object B with the same charge as A is placed at 0.250m south of A (so that objects A and B and point P follow a straight line), what is the magnitude of the total electric field produced by the two objects at P? I know that E = (KQ)/r^2 so E = (KQ)/r^2 E = ((9×10^9)(2.78×10^−10)/0.5^2 E = 10 So the total magnitude produced by the two objects 10 + 40 = 50N/C?
Correct.
 P: 452 Got both questions right. Thank you all.

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