# Homework Help: Find the electric field

1. Mar 30, 2008

### ~christina~

1. The problem statement, all variables and given/known data
A charge $$q_1= 7.0\mu C$$ is located at the origin, and a second charge $$q_2= -5.0 \mu C$$is located on the x-axis, 0.30m from the origin. Find the electric field (maginitude and direction) at the point P, which has the coordinates (0,0.40)m.

2. Relevant equations
$$E = k_e \sum \frac{q} {r_i^2} \hat{r}$$

3. The attempt at a solution

well,

$$q_1= 7.0 \mu C$$

$$q_2= -5.0 \mu C$$

r= 0.30m

y= 0.40m

how would I find the field if I'm given the coordinates of 0, and 0.40 which would mean that If I'm reading that right. then it would mean that it's 0.40m above the charges in the y direction?

Not sure if this picture is correct but I tried to draw the layout of the problem.

Last edited: Mar 30, 2008
2. Mar 30, 2008

### nrqed

yes, it looks like you interpreted the question correctly (although I personally would have thought that q2 would be to the right of q1 but the question is ambiguous about this point).

Find th emagnitude and direction of the Efield produced by each charge at that point, then decompose them into x and y components and find the total Efield.

By the way, I am curious: what software did you use to make this nice drawing?

3. Mar 30, 2008

### ~christina~

I just noticed that I wrote, $$q= -4 \mu$$ when it's supposed to be $$q= -5 \mu$$ and I put it on that side since I thought that since it was negative it should be on the left side but thinking about it again it shouldn't really matter since it's a charge I think.

but if the charge is not at an angle but right below the point P would the r be just 0.40m?

$$E_1= k_e \frac{|q_1|} {r_1^2}= 8.9876x10^9N*m^2/C^2 \frac{|7.0 \mu C|} {0.40m^2}= = 3.932 x 10^{11}$$

and

$$E_2 = 8.9876x10^9N*m^2/C^2 \frac{|-5.0\mu C|} {0.30^2+ 0.40^2}= 3.12x10^{12}$$

(not sure the units of the Electric field either)

for each it woud be

$$E_2 = 8.9876x10^9N*m^2/C^2 \frac{|-5.0\mu C|} {0.30^2+ 0.40^2}= 3.12x10^{12} cos \hat{i} - 3.12x10^{12} sin \hat{j}$$

and

$$E_1= 8.9876x10^9N*m^2/C^2 \frac{|7.0\mu C|} {0.30^2+ 0.40^2}= 3.932 x 10^{11} cos \hat{i} + 3.932 x 10^{11} sin \hat{j}$$

not sure how to divide the forces for the first q1 though
$$E_x= 3.12x10^{12} cos (45) j + 3.932 x 10^{11} cos (90)j$$

$$E_y= 3.12x10^{12} sin (45) j + 3.932 x 10^{11} sin (90)j$$

Is this right?

nice software? => paint (thus no theta sign)

4. Mar 30, 2008

### nrqed

The charges are in microcoulombs. One microcoulomb is how many coulombs?
You should be able to tell by simply including the units on all the quantities in your equations and simpifying the units.
I guess you forgot to include the angles in the trig functions! If you give the angle with respect to the positive x axis, all the terms should be positive. But some people prefer to use angles defined in a different way (from a different axis) so it varies. I can't tell you if your signs are correct if you don't tell me what are your angles.

No. You are mixing components with vectors. A component is NOT a vector so what you wrote does no make sense. You should not have put i and j unit vectors there.

And how did you decide that one of the angles was 45 degrees? This is not the case!

5. Mar 30, 2008

### ~christina~

$$1C= 1x10^-6 \mu C$$

okay. Is it N/m*C ?

hm..well to find the angles I guess I have to do trig. (it looked like 45 so I used 45)
so. 0.30^2 + 0.40^2= 0.5 (hypotenuse)
$$sin ^{-1}(0.40m / 0.50m)= 53.13^o$$

$$E_2 = 8.9876x10^9N*m^2/C^2 \frac{|-5x10^{-6} C|} {0.30^2+ 0.40^2}= -7.19x10^{5} cos(53.13) \hat{i} - (-7.19x10^5) sin (53.13)\hat{j}$$

and

$$E_1= 8.9876x10^9N*m^2/C^2 \frac{|7x10^{-6}C|} {0.30^2+ 0.40^2}= 2.51x10^5 cos (90)\hat{i} + 2.51x10^5 sin(90) \hat{j}$$

is this right?

So it's right if it has no vectors?

Last edited: Mar 30, 2008
6. Mar 31, 2008

### nrqed

there should only be one minus sign in the second term (the y component of the E field should come out negative, right?)
Probably a typo but you need to divide by 0.40^2 only.

I did not check the actual numerical values but it looks good.

7. Apr 5, 2008

### ~christina~

$$E_2 = 8.9876x10^9N*m^2/C^2 \frac{|-5x10^{-6} C|} {0.30^2+ 0.40^2}= -7.19x10^{5} cos(53.13) \hat{i} - (-7.19x10^5) sin (53.13)\hat{j}$$

oh yep I typed that before but it should be only 0.40^2

$$E_1= 8.9876x10^9N*m^2/C^2 \frac{|7x10^{-6}C|} {0.40^2}= 2.51x10^5 sin(90) \hat{j}$$

alright is this fine now? I think I would add them together right?

8. Apr 5, 2008

### nrqed

Looks fine (except that there is still two minus signs in the y component of $$\vec{E}_1$$; there should be only one minus sign).

yes, add up all the x and y components.

9. Apr 5, 2008

### ~christina~

$$E_2 = 8.9876x10^9N*m^2/C^2 \frac{|-5x10^{-6} C|} {0.30^2+ 0.40^2}= -7.19x10^{5} cos(53.13) \hat{i} + (-7.19x10^5) sin (53.13)\hat{j}$$

I thought you meant E2 but I fixed E1 below...right ? E 1

$$E_1= 8.9876x10^9N*m^2/C^2 \frac{|7x10^{-6}C|} {0.30^2+ 0.40^2}= 2.51x10^5 sin(90) \hat{j}$$

and I think I shouldn't have a x component for E1 since it is only directed in the y direction...which is up thus possitive

I'm not sure what you are speaking of unfortunately (if the above is incorrrect)

10. Apr 7, 2008

### ~christina~

Correction from above

$$E_2 = 8.9876x10^9N*m^2/C^2 \frac{|-5x10^{-6} C|} {0.30^2+ 0.40^2}= -7.19x10^{5} cos(53.13) \hat{i} + (-7.19x10^5) sin (53.13)\hat{j}$$

I thought you meant E2 but I fixed E1 below...right ? E 1

$$E_1= 8.9876x10^9N*m^2/C^2 \frac{|7x10^{-6}C|} {0.40^2}= 2.51x10^5 sin(90) \hat{j}$$

and I think I shouldn't have a x component for E1 since it is only directed in the y direction...which is up thus possitive

I'm not sure what you are speaking of unfortunately (if the above is incorrrect)