Magnitude and Direction of Electric Fields

In summary, the charge of object A, which is creating an electric field at a point P located north of A, is determined to be 2.78x10^-10C. The charge is negative due to the direction of the electric field pointing towards the charge. If a second object B with the same charge as A is placed south of A, the total magnitude of the electric field at point P is 50N/C.
  • #1
cse63146
452
0

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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  • #2
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:
  • #3
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
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.
 
  • #5
so it would be positive?
 
  • #6
cse63146 said:
so it would be positive?

Why would you think that?
 
  • #7
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
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
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.
 

1. What is the definition of electric field magnitude?

The magnitude of an electric field at a specific point is the force per unit charge that would be experienced by a positive test charge placed at that point.

2. How is the magnitude of an electric field calculated?

The magnitude of an electric field can be calculated using the equation E = F/q, where E is the electric field, F is the force, and q is the charge of the test charge.

3. What factors affect the magnitude of an electric field?

The magnitude of an electric field is affected by the distance between charges, the amount of charge on the source, and the medium through which the electric field is passing.

4. How is the direction of an electric field determined?

The direction of an electric field is determined by the direction in which a positive test charge would be pushed or pulled. It is always directed away from positively charged sources and towards negatively charged sources.

5. What is the relationship between electric field magnitude and electric potential?

Electric potential is directly related to electric field magnitude. The higher the electric field magnitude, the greater the change in electric potential over a given distance.

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