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 Question Magnitude and Direction of Electric Fields ( Physics Help and Math Help - Physics Forums Introductory Physics )Updated: 2008-03-11 17:07:06 (12)
 Magnitude and Direction of Electric Fields 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?
 Answers: Magnitude and Direction of Electric Fields ( Physics Help and Math Help - Physics Forums Introductory Physics )
Magnitude and Direction of Electric Fields

 Originally Posted 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.)

dynamicsolo

Magnitude and Direction of Electric Fields

 Originally Posted 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.)

dynamicsolo

 Magnitude and Direction of Electric Fields 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. belliott4488
Magnitude and Direction of Electric Fields

 Originally Posted 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.

dynamicsolo

 Magnitude and Direction of Electric Fields so it would be positive? cse63146
Magnitude and Direction of Electric Fields

 Originally Posted by cse63146 so it would be positive?
Why would you think that?

Dick

 Magnitude and Direction of Electric Fields 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? belliott4488
 Magnitude and Direction of Electric Fields 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? cse63146
Magnitude and Direction of Electric Fields

 Originally Posted 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.

dynamicsolo

 Magnitude and Direction of Electric Fields 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? cse63146
Magnitude and Direction of Electric Fields

 Originally Posted 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.

elecstorm

 Magnitude and Direction of Electric Fields Got both questions right. Thank you all. cse63146
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