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Old Jan 11th 2015, 09:40 AM   #1
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Electric field problem

Hi all the members,

Since this is my first posting here this year, I would like to wish you first a Happy New 2015 Year!!! to all of you

Now, the problem:
Four charged particles are at the corners of a square of side a(I will try to show it - the right side is pushde to the left when I save the square???). (Let A = 5, B = 5, and C = 5.)

qA-------q
| |
| |
| |
qB-------qC

a) I need to determine the electric field at the location of charge q. (Use the following as necessary: q, a, and ke.)

so, since the electric field at charge q will be the sum of the electric fields caused by Aq, Bq and Cq I take it one by one:

for Aq, E= k_e*(5q/a^2)i + 0j
for Bq, E= k_e*(5q/a^2)[a/sqrt(2a^2)]i + k_e*(5q/a^2)[a/sqrt(2a^2)]j =
= k_e*(5q/a^2)[1/sqrt(2)]i + k_e*(5q/a^2)[1/sqrt(2)]j
and for Cq, E = 0i + k_e*(5q/a^2)j

now, if we sum them all and take the common factor, we get
E_total = k_e*(5q/a^2)i + k_e*(5q/a^2)[1/sqrt(2)]i + k_e*(5q/a^2)[1/sqrt(2)]j + k_e*(5q/a^2)j =
= k_e*(5q/a^2){1i + [1/sqrt(2)]i + [1/sqrt(2)]j +1j}=
= k_e*(5q/a^2){1.707i + 1.707j} = k_e*(q/a^2){12.07} N/C, the magnitude of the electric field

Can you please check my work, because my answer is considered as incorrect, and I have no idea why

Thanks

Last edited by dokrbb; Jan 11th 2015 at 09:43 AM. Reason: the square doesn't keep the right side in place
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Old Jan 11th 2015, 12:34 PM   #2
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Never mind; I got it, the mistake was when I was calculating the unit vector for the charge at position B
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