Magnitude and Direction of Electric Fields


by cse63146
Tags: direction, electric, fields, magnitude
cse63146
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#1
Jan11-08, 09:37 PM
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) 1.1110^−9C
B) −1.1110^−9C
C) 2.7810^−10C
D) −2.7810^−10C
E) 5.7510^12C
F) −5.7510^12C


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))/910^9
= 2.7810^−10C

makes sense to me, but should it be negative or positive?
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dynamicsolo
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#2
Jan11-08, 10:16 PM
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Quote Quote by cse63146 View Post
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.)
belliott4488
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#3
Jan11-08, 10:20 PM
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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.

dynamicsolo
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#4
Jan11-08, 10:24 PM
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Magnitude and Direction of Electric Fields


Quote Quote by belliott4488 View Post
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.
cse63146
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#5
Jan11-08, 11:57 PM
P: 452
so it would be positive?
Dick
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#6
Jan12-08, 12:21 AM
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Quote Quote by cse63146 View Post
so it would be positive?
Why would you think that?
belliott4488
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#7
Jan12-08, 07:53 AM
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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?
cse63146
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#8
Jan12-08, 09:28 AM
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?
dynamicsolo
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#9
Jan12-08, 11:58 AM
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Quote Quote by cse63146 View Post
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.
cse63146
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#10
Jan12-08, 12:25 PM
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 = ((910^9)(2.7810^−10)/0.5^2
E = 10

So the total magnitude produced by the two objects 10 + 40 = 50N/C?
elecstorm
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#11
Jan12-08, 12:34 PM
P: 3
Quote Quote by cse63146 View Post
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 = ((910^9)(2.7810^−10)/0.5^2
E = 10

So the total magnitude produced by the two objects 10 + 40 = 50N/C?
Correct.
cse63146
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#12
Jan12-08, 12:42 PM
P: 452
Got both questions right. Thank you all.


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