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## Homework Statement

Point charge q1 = -5.0nC is at the origin and point charge q2 = +3nC is 3cm in the +x direction. Point P is at y = 4cm. (a) Calculate the electric fields E1 and E2 at P due to charges q1 and q2, expressed in unit vector notation. (b) Use the results from the previous question to find the resultant electric field E at P, expressed in unit vector form.

## Homework Equations

$$ \vec{E} = \frac{1}{4\pi\epsilon_{0}} \frac{q}{r^{2}}\hat{r} $$

## The Attempt at a Solution

Simply plugging in the constants (q1 = -5 x 10^-9, q2 = 3 x 10^-9, r1 = .04, r2 = .05) gets the following values for E1 and E2:

|E1| = -28125 V

|E2| = 10800 V

Based on this, I expected the magnitude of the resultant electric field to be -17325 V.

Then, to solve using unit vectors, I drew the components from point P:

The vector E1 is raised 36.9 degrees west of north, which would give the components of E1 as follows:

E1 = <-10800sin(36.9), 10800cos(36.9)>

Since E2 is simply pointed down toward the origin, it has an angle of zero, leaving it as:

E2 = <0, -28125>

And adding the two up for the vector sum:

E = <-10800sin(36.9) + 0, 10800cos(36.9)-28125>

So far this seemed entirely correct. But when I checked the calculation by determining the magnitude of the sum, it came out to about 20,500 V. I figure I must be missing something obvious, but I can't quite figure out what it is.