Physics Help Forum Electric field problem

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 Jan 11th 2015, 09:40 AM #1 Member     Join Date: Oct 2013 Posts: 48 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
 Jan 11th 2015, 12:34 PM #2 Member     Join Date: Oct 2013 Posts: 48 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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