Magnitude of electric field using point charges

In summary, the magnitude of the electric field at point P, located at (5.45 cm, 0), due to Q1 alone is 7.56x10^6 N/C.
  • #1
n77ler
89
0
[SOLVED] Magnitude of electric field using point charges

Homework Statement


Two charges, Q1 = 3.15 μC and Q2 = 5.40 μC, are located at points (0, -2.80 cm) and (0, +2.80 cm), What is the magnitude of the electric field at point P, located at (5.45 cm, 0), due to Q1 alone?


lQ2
l
l
l
l--------------P
l
l
l
l
lQ1



Homework Equations





The Attempt at a Solution

 
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  • #2
In order to get help, post your attempted solution and point out where you got stuck.
 
  • #3
I used E=( k lq1l ) / r^2

(8.99x10^9) (3.15x10^-6) / (0.0612m)^2
=7.56x10^6 N/C
 
  • #4
n77ler said:
I used E=( k lq1l ) / r^2

(8.99x10^9) (3.15x10^-6) / (0.0612m)^2
=7.56x10^6 N/C
Looks good to me.
 
  • #5
k got it, I think I was putting in 10^-6 by accident :S lol, thanks!
 
  • #6
Ok so there were two other parts to the question, I had to find the xcomponent and ycomponent of total electric field at P. I solved the one for x and got it correct which means I must be using the right angle and stuff to do my calculations. My answer for y was wrong, why is this?
 
  • #7
n77ler said:
My answer for y was wrong, why is this?
Beats me. Show what you did.
 
  • #8
Workings are on attached file...
 

Attachments

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  • #9
Check your signs. Realize that one field points up, while the other points down.
 
  • #10
opps so there should be subtraction to get my final answer...
So the Q1 diagram needs to be a negative answer...

(-3.44x10^6)+ (5.90x10^6)
=2.46x10^6N/C
Thats still not right tho, I don't know what I am going wrong
 
  • #11
n77ler said:
opps so there should be subtraction to get my final answer...
So the Q1 diagram needs to be a negative answer...
Q1 is the one on the bottom, right? So where does its field point?
 
  • #12
Ok, so the Q1 line points in the positive direction while the Q2 points in the negative direction right? I got the right answer but I just want to make sure my reasoning for sign convention is understood.
 
  • #13
Right. The field from a positive charge always points away from the charge. So the y-component of the Q1 field is + and the Q2 field is -.
 
  • #14
Ok thanks once again!
 

1. What is the equation for calculating the magnitude of electric field using point charges?

The equation for calculating the magnitude of electric field using point charges is given by E = (kQ)/r^2, where E is the magnitude of electric field, k is the Coulomb's constant (9x10^9 N*m^2/C^2), Q is the charge of the point charge, and r is the distance between the point charge and the location where the electric field is being measured.

2. How do I determine the direction of the electric field using point charges?

The direction of the electric field is determined by the direction of the force that the electric field exerts on a positive test charge placed at a particular location. The direction of the electric field is always away from positive charges and towards negative charges.

3. Can the magnitude of electric field be negative?

Yes, the magnitude of electric field can be negative. This occurs when the point charges have opposite signs and the electric field at a particular location is directed towards the negative charge instead of away from it.

4. How does the distance between point charges affect the magnitude of electric field?

The magnitude of electric field is inversely proportional to the square of the distance between point charges. This means that as the distance between point charges increases, the magnitude of electric field decreases.

5. What is the unit of measurement for the magnitude of electric field?

The unit of measurement for the magnitude of electric field is Newtons per Coulomb (N/C) or Volts per meter (V/m).

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