# Stuck on two electric field questions

1. Jul 12, 2005

### DAP1MP13

Stuck on two electric field questions....

Hi guys,
I have these two questions which I am unable to answer.

This is the first one:

Calculate the magnitude of the electric field at one corner of a square 1.26 m on a side if the other three corners are occupied by 2.80×10-6 C charges .

Hint: Draw the diagonal through the square from the lower left with one of the 3 charges to the upper right with the point where the electric field is wanted; the resulting field due to the 2 other source charges on either side of the diagonal points along the diagonal (why ?); calculate the magnitude for the vector sum; add this value to the field due to the charge on the lower left which also points along the diagonal (simply add the numbers).

I don't really even know how to start....(there is an example like this in my text book 'physics'by Giancoli, but it is quite different from this question)

the second one:
A positive point charge Q1 = 3.43×10-5 C is fixed at the origin of coordinates, and a negative charge Q2 = -7.27×10-6 C is fixed to the x axis at x = +1.97 m. Find the location of the place along the x axis where the electric field due to these two charges is zero.

E = kQ/r^2
I set up E = 0 = 308700/1.97^2 - 65430/1.97^2

I set it to equal zero, but obviously that is incorrect.
I think I'm getting confused because one of the charges is set at the origin, it threw me off a little bit. So I know that for the first part the "r" should not be 1.97....

Any help is appreciated.... thanks....

Last edited: Jul 12, 2005
2. Jul 12, 2005

### DAP1MP13

for the second question I tried setting this up....

308700/x^2 = 65430/1.97^2
so x = 4.28m

But that is not right either....
I don't know what to do....

3. Jul 12, 2005

### Pyrrhus

For the first problem you simply draw the vector acting on the corner and add them like vectors, remember each vector modulus or magnitudes can be found throught $E = k_{e} \frac{|q|}{r^2}$

4. Jul 13, 2005

### Pyrrhus

For the second problem, you can try to guess where the that point will be, it can't be between the charges, because the resultant electrical field won't be 0, both vectors are pointing in the same way, so it must be outside from there. You can try it on the right, suppose it's a distance x from the charge away from the origin, so the r for the first charge will be 1.97 + x, while for the second charge will be x.

5. Jul 13, 2005

### DAP1MP13

ok, so I got the hypotnuse of the square to be 1.78191m.
After that I used the eqution given above.

E=(9e9)*2.80e-6/1.78191^2
=7936.5 N/C

I muliplied that by three to get 23809.5 N/C, and that is not the correct answer either. I tried the problem again, but I used the side of the square (1.26m) as "r", that is wrong also. I know this problem isn't that hard, I'm just starting to get frustrated....
Anymore hints would be great.

Also, I wouldn't need to subtract and thing? right? Since all three charges and equal and positive?

Thanks for the help so far.

6. Jul 13, 2005

### Pyrrhus

Are you familiar with unit vectors?, well for rectangular systems such as this the ones used are $\vec{i}, \vec{j}, \vec{k}$

I will set up the rectangular components for each electric field vector.

$$\vec{E}_{1} = k_{e} \frac{|q|}{l^2} \vec{i}$$

$$\vec{E}_{2} = -k_{e} \frac{|q|}{l^2} \vec{j}$$

$$\vec{E}_{3} = k_{e} \frac{|q|}{l^2 + l^2} \frac{l}{\sqrt{l^2 + l^2}} \vec{i} - k_{e} \frac{|q|}{l^2 + l^2} \frac{l}{\sqrt{l^2 + l^2}} \vec{j}$$

7. Jul 13, 2005

### DAP1MP13

Cyclovenom, yes I am familiar with unit vectors.
Wow, I was oversimplifying the problem. I feel kinda dumb for not thinking straight.
I will work on the problem tomorrow morning.

thank you for the great help once again!

8. Jul 13, 2005

### Pyrrhus

No problem, and Welcome to PF!

9. Jul 13, 2005

### DAP1MP13

Cyclovenom

I couldn't go to sleep before trying it again, haha.
But I got the first one correct, finally.

I ended up doing this E = kQ/((2/L^2)*cos45+(1/hyp^2))
So as the final answer I got 30384.3 N/C!

Still confused like crazy about the second question though.....

10. Jul 13, 2005

### Pyrrhus

2nd problem start setting the electric field vectors.

$$\vec{E}_{1} = k_{e} \frac{|q_{1}|}{(x+1.97)^{2}}\vec{i}$$

$$\vec{E}_{2} = - k_{e} \frac{|q_{2}|}{x^{2}}\vec{i}$$

where

$$\vec{E}_{1} + \vec{E}_{2} = 0$$