E Field Magnitude at Origin due to Single Charge 5cm Above

In summary, the conversation discusses the calculation of the magnitude of the electric field at the origin due to a single charge located 5 cm above the horizontal axis. While the earlier question asked for the direction and magnitude of the field due to a single negative charge on the horizontal axis, the present question allows for either a positive or negative charge. Through the use of Coulomb's law, it is determined that the magnitude of the electric field can vary depending on the exact placement of the charge, as seen in the two given examples. The conversation ends with the clarification that the charge is not located directly above the origin, but rather off in the corner, leading to a different distance from the origin and ultimately a different magnitude of the electric field.
  • #1
sasuke07
54
0

Homework Statement


What is the magnitude of the E field at the origin due to a single charge (either the + or the -) located 5 cm above the horizontal axis?

Homework Equations


KQ/D^2



The Attempt at a Solution


in an earlier question it asked What is the direction and magnitude of the E field at the origin due to the single negative charge (ignore all others) on the horizontal axis?" SO i used columbs law "KQ/d^2, (8.99x10^9x30x10^-9)/.05^2 and i got 1.1x10^5 which was the correct answer, but this questions answer is supposed to be 5.4X10^4. I was wondering why this is the case as all the numbers are the same from an earlier question, exepct now its vertical.
 

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  • #2
Please provide earlier question :)
 
  • #3
the earlier question was "What is the direction and magnitude of the E field at the origin due to the single negative charge (ignore all others) on the horizontal axis?"
 
  • #4
and the present question?
 
  • #5
I didn't get how all the numbers are same in both question , aren't they different
 
  • #6
Because in both questions you are still 5 cm away from the e field, and the charges are still 30 nc. The first question the negative charge is to the left of the center and in the second question the charge (the problem said could be + or -) is 5 cm above the center. So aren't the numbers the same if i were to use the same equation?
 
  • #7
Sorry in question 1 the negative charge is 5 cm to the left of the origin and question 2 the charge is 5 cm above the origin.
 
  • #8
but the number of charges are different , they are infinite
and also they are distributed along lines not at a point
 
  • #9
just for the sake of imagination
think of a person pushing a box with the help of a rope alone
and think many people pushing a box with many ropes
 
  • #10
aren't there only 1 charge in both questions as question 1 says to ignore the other charges and question 2 says there is a single charge? would i have to use a different formula?
 
  • #11
i really wish if you could post both the question exactly separated by space , i am confused
 
  • #12
Sorry
1."What is the direction and magnitude of the E field at the origin due to the single negative charge (ignore all others) on the horizontal axis?"

2.What is the magnitude of the E field at the origin due to a single charge (either the + or the -) located 5 cm above the horizontal axis?

I know how to do 1, its just question 2 that's tripping me up.
 
  • #13
where is the particle placed in Q1
 
  • #14
particle is at the origin, and the negative charge is on the left
 
  • #15
left ? what is the coordinate?
 
  • #16
If you look at the picture the particle would be at the center and the negative charge is on the x-axis to the left of the origin. So it would be 5 cm away, as the entire horizantel axis is 10cm.
 
  • #17
see magnitude of electric field will be same for all points whose distance from the particle is same
 
  • #18
locus will be a sphere , where the whole surface will have same magnitude at all points on the surface
 
  • #19
See magnitude of electric field is same for all points which are equidistant to the particle ( which is a locus of a hollow sphere )
 
  • #20
Thats what i was thinking, but on the answer sheet the it said the magnitude of
1. was 1.1x10^5 N/C to left and for 2 it said the magnitude was 5.4X10^4 N/C. I am stumped as to how the magnitude could be different when all the numbers used is essentially the same.
 
  • #21
There might be something else you are missing
start completely new :0
 
  • #22
would you have an idea of what i could be missing. I posted everything related to the 2 questions and i can't think of anything. The only formula you could use is Columbs law, so ya
 
  • #23
See i can only think if you write both the questions exactly as they are without omitting anything , with the figure and everything given
 
  • #24
Im uploading a copy of where i am getting the questions. on the quiz they are questions 4 and 7. And you have the picture from my first post. Thats all the information that i received for this assignment.
 

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  • #25
questions 7 and 8 are the only 2 that i can't figure out for myself. So if you could help me out with either of the 2 that would be awesome.
 
  • #26
lol , this is so stupid ,
ok see here you are
In the 7th Question they are asking electric field due to a charge located 5 cm above horizontal axis in that given figure only .
now there can be 2 possible cases
1. + q (5,5)
2. - q (-5,5)

since they have asked for only one you take anyone of them
and the distance from the origin won't be 5
why? tell me
 
  • #27
holy ****, is it because the charge isn't on the vertical axis right above the origin but its off in the corner.
 
  • #28
Yea they haven't mention anything about x coordinate
and even in the figure there is no charge at (0.5)
There you go .

keep visiting PF :)
 
  • #29
thanks for sticking with this. can't believe i didn't see that.
 

FAQ: E Field Magnitude at Origin due to Single Charge 5cm Above

1. What is the formula for calculating the E field magnitude at the origin due to a single charge 5cm above?

The formula for calculating the E field magnitude at the origin due to a single charge 5cm above is E = k * (q / r^2), where k is the Coulomb's constant (9 * 10^9 N.m^2/C^2), q is the magnitude of the charge, and r is the distance between the charge and the origin.

2. How do I determine the direction of the E field at the origin?

The direction of the E field at the origin is determined by the direction of the force on a positive test charge placed at the origin. If the charge is positive, the direction of the E field is away from the source charge, and if the charge is negative, the direction of the E field is towards the source charge.

3. Can the E field magnitude at the origin be negative?

Yes, the E field magnitude at the origin can be negative if the source charge is negative. This means that the direction of the E field is towards the source charge instead of away from it.

4. How does the E field magnitude at the origin change as the distance between the charge and the origin increases?

The E field magnitude at the origin decreases as the distance between the charge and the origin increases. This is because the strength of the electric field decreases as the distance from the source charge increases according to the inverse square law.

5. What is the unit of measurement for the E field magnitude at the origin?

The unit of measurement for the E field magnitude at the origin is Newtons per Coulomb (N/C). This unit represents the amount of force per unit charge at a certain point in space.

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