1. Feb 9, 2009

NicoleKarmo

1. The problem statement, all variables and given/known data
Find the electric field and the potential at the center of a square 2.0 cm with charges of +9.0uC at one corner of the square and with charger of -3.0uC at the remaining 3 corners.

2. Relevant equations
the sum of Vi= the sum of KQi/ri

3. The attempt at a solution

I saw that the equation was re-arranged this way but I'm not sure how they arrived there:
E= Ea+Eb+Ec+Ed = Ea+Eb+Ec-Eb= Ea+Ec= (K|qa|/r^2 + K|qc|/r^2) towards C

when you plug in the #'s the answer is 5.4x10^8 N/C toward c.
what I dont understand is how they concluded that that was the formula to use.

2. Feb 9, 2009

LowlyPion

Welcome to PF.

The E-field is a vector field.

The scalar form of the equation can be used to calculate the magnitude of the field at a point. But when you have multiple charges, then you have to pay attention to the direction of all the charges.

When calculating the field at the center then, the opposite corners of the square will act oppositely. Hence when you compare them, the corners with the same charges will offset (equal and like charges equidistant in opposite directions and all that).

That just means then that you need only calculate the magnitude of the different charges because ... well they will make different contributions.