# Finding resultant electric field

1. Jan 27, 2014

### cwbullivant

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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.

2. Jan 27, 2014

### hjelmgart

Uhm, where is the point, P located in the x-direction? How did you calculate r1 and r2 with respect to that?

To me it seems like you need to do it component wise. So you need to seperate
$$\hat{r}$$
into two components, x and y and give it for instance
$$\hat{x}$$
and
$$\hat{y}$$
for unit vectors. Are you unfamiliar with this kind of notation?

Now you have to calculate the electric field in the x and y direction seperately. Say you want to calculate the electric field E3 at a point at x=3 and y=5, you would then use the equation in the following manner:

$$E3 = \frac{1}{4\pi \epsilon_0} \frac{q}{3\hat{x} + 5\hat{y}}$$

Last edited: Jan 27, 2014
3. Jan 27, 2014

### cwbullivant

Sorry, P is located at 0 in the x direction. The distances were calculated by converting them to meters, and the distance from q2 to P was done by noting that the right triangle created by q1q2P is a Pythagorean triple (3,4,5, in this case).

4. Jan 27, 2014

### lightgrav

2) you don't add the x-component to the y-component. keep the 6480 V/m (x) separate!

5. Jan 27, 2014

### cwbullivant

1) Noted. I forgot that Potential and Electric field have different units.

2) I didn't add the x-component to the y-component, as far as I can see. The sin(36.9) is associated with the x-component of E2 because the angle is measured from the y-axis (I think this might be why it looks like that).

E1 = 0(i) + 28125(-j)

E2 = 10800sin(36.9)(-i) + 10800cos(36.9)j

(I mixed up E1 and E2 in the original post, but the components were still the same)

6. Jan 28, 2014

### hjelmgart

For starters I would believe, you swapped the sine and cosine? So cosine should relate r to the x-direction, and sine would relate r to the y direction, though I doubt it changes much.

Also the way you "check" your result seems to be odd. When you add E1 and E2 you get a smaller magnitude for the resulting electric field. Does that make sense to you?
You have two charges of opposite charges, what would you usually expect happens to the electric field in such case? Would they attract or repel, and thus either increase or decrease at a point between them?
A drawing is always good for this.

Besides of that the math seems to be correct, and if I check it with the numbers you provided, I do get the same results in both ways that you did.

Last edited: Jan 28, 2014