Magnitude and Direction of Electric Fields

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Homework Help Overview

The discussion revolves around determining the charge of an electrically charged object A based on the electric field it produces at a point P. The electric field at point P is given as 40.0 N/C directed south, with the distance from A to P being 0.250 m. The participants explore the implications of the electric field's direction on the sign of the charge.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the electric field direction and the sign of the charge, questioning whether the charge should be positive or negative based on the field's behavior. Some participants reference the definition of electric fields in relation to test charges.

Discussion Status

The discussion is active, with participants exploring the implications of the electric field's direction on the charge's sign. Some guidance has been offered regarding the relationship between the electric field and the nature of the charge, but no consensus has been reached on the final determination of the charge's sign.

Contextual Notes

Participants note that the problem includes a second part regarding the total electric field produced by two charges, which adds complexity to the discussion. The implications of the charge's sign on the total electric field are also being considered.

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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?
 
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cse63146 said:
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:
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 said:
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.
 
so it would be positive?
 
cse63146 said:
so it would be positive?

Why would you think that?
 
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?
 
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 said:
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
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
cse63146 said:
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
Got both questions right. Thank you all.
 

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