Homework Help: Finding the electric field at 3 points due to 2 identical charges.

1. Jan 22, 2012

get_physical

3 points (a,b and c) and two identical positive charges. How do you find MAGNITUDE of the electric field at the 3 points? I just need to rank them, don't need numbers.

all points are on the x-axis:
A (x=0); charge +Q(x=1); B (x=3); c(x=4); charge+Q(x=5)

(Picture attached in the 3rd post)

2. Relevant equations

E= (1/4pie) (q/r^2)

3. The attempt at a solution
I just focused on the r^2 since everything in the formula is the same.
for point A: 1^2+ 5^2 = 26
Point B: 2^2+2^2 = 8
Point C: 1^2 + 3+2 = 10

so would B then have the greatest strength while A has the least?

Last edited: Jan 22, 2012
2. Jan 22, 2012

SammyS

Staff Emeritus
r2 is in the denominator. You need to add or subtract 1/12 and 1/52, 1/22 and 1/22, etc.

Also you haven't taken into account the direction of the E field from each charge.

3. Jan 22, 2012

get_physical

Picture attached

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4. Jan 22, 2012

get_physical

Yes, r^2 is the denominator, that's why the smallest r^2 will have the strongest field. Can you please explain what you mean I need to add or subtract?

5. Jan 22, 2012

get_physical

In that case, would there be no electric field strength at B? since the directions are opposite, so they cancel each other out?

6. Jan 22, 2012

SammyS

Staff Emeritus
Yes.

As for using 1/r, not r2:
1/1+1/52 = 25/25 + 1/25 = 26/25

1/22 + 1/22 = 1/4 + 1/4 = 1/2   (Yes, I know these should be subtracted, but this is just to illustrate the point.)​

7. Jan 22, 2012

get_physical

In this case, after doing the calculations, would a have the smallest electric field strength since it is -26/25, and then after that would be B with 0 and finally, C?

Last edited: Jan 22, 2012
8. Jan 22, 2012

get_physical

would you just assume the left is negative while if arrows pointing to the right is positive?
Thanks

9. Jan 22, 2012

SammyS

Staff Emeritus
What do you get for electric field at point C ?

Look at the problem statement. It asks about the "MAGNITUDE of the electric field". (That capitalization was in your original post.)

As with number lines, it's fairly standard to consider a vector pointing to the right to be positive, and left is negative.

10. Jan 22, 2012

get_physical

Oh !!! thank you so much I finally got it. thank you for your patience. appreciated!