(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

One side of a square has length 2.0 cm.

In three of the square's corners, there are point charges.

Top left corner: -1.0 nC (call this A)

Bottom left corner: 10 nC (call this B)

Bottom right corner: -1.0 nC (call this C)

The top right corner has no charge, and is labelled P.

What is the magnitude of the electric field at point P?

2. Relevant equations

Electric Field = KQ/r^{2}

3. The attempt at a solution

Electric Field Due to A:

K(-1.0 x 10^{-9}C)/(0.02 m^{2}) = -22475 N/C, left

Electric Field Due to C:

Same as above, but this time the direction is down

Electric Field Due to B:

First calculate the distance from B to P. This is 0.02^{2}+0.02^{2}= 0.0008. Take the square root of 0.0008 to find 0.0282 m.

K(10 x 10^{-9}C) / (0.0282)^{2}= 112375 N/C, at a 45 degree angle pointing north east.

Now, take the sum of all three parts to find the net electric field.

A and C together point south west, with a magnitude of 31784.44 N/C.

112375 N/C - 31784.44 N/C = 80590.55 N/C, north east.

However, the answer given on the page is not what I got. What went wrong?

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# Homework Help: Calculate the Electric Field at One Corner of a Square

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