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Magnitude and Direction of Electric Fields

  • Thread starter cse63146
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  • #1
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Homework Statement



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?

Homework Equations



E = (KQ)/r^2

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 and Replies

  • #2
dynamicsolo
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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.)
 
Last edited:
  • #3
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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.
 
  • #4
dynamicsolo
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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.
 
  • #5
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so it would be positive?
 
  • #6
Dick
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so it would be positive?
Why would you think that?
 
  • #7
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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?
 
  • #8
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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?
 
  • #9
dynamicsolo
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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.
 
  • #10
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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?
 
  • #11
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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.
 
  • #12
452
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Got both questions right. Thank you all.
 

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