# Electric Field Q

1. Aug 7, 2007

### t_n_p

The problem statement, all variables and given/known data

Two charges -1nC and +1nC are positioned at (-2,0) and (2,0) respectively, where distances are in cm.

a) Sketch 5-6 field lines E in the region bounded by x=-2, x=2, y=0 and y=8

b) Caclculate the horizontal and verticle components of the E field at (-2,6), Draw a diagram showing the charges, the fields at P due to both charges and the resultant field. Find magnitude E and the angle phi (measured from the +x-axis)

The attempt at a solution

a)http://img241.imageshack.us/img241/1310/untitledxg6.jpg [Broken]

b) I'm not sure how to calculate the hotizontal and ver components of the E at (-2,6)!

Last edited by a moderator: May 3, 2017
2. Aug 7, 2007

### Staff: Mentor

What's the electric field at that point due to each point charge? Add the two fields--they are vectors--to get the net field at that point. (Hint: Find and add the components.)

3. Aug 7, 2007

### t_n_p

Do I use coulombs law, |F| = (k)(|q1|)(|q2|)/(r^2)

and then E = F/Qo ?

I found the second formula in my notes, but not sure how Qo comes into it?

4. Aug 7, 2007

### Staff: Mentor

Yes.

Let q1 be your point charge (for which you are finding the field) and q2 be a test charge at point P. q2 = Qo.

When all is said and done, the magnitude of the field from a point charge Q is:
|F| = k|Q|/(r^2)

5. Aug 8, 2007

### t_n_p

P doesn't appear to have a charge, so would I use
|F| = (k)(|q1|)(|q2|)/(r²)
or
|F| = k|Q|/(r²)?

6. Aug 10, 2007

### t_n_p

I think I pretty much managed to do it..
Basically I used formula E = kq/r2 to find the E on P by point A(2,0) and the force on P by point B(-2,0).

Then I found the x and y components of the E on P by A.
Then calculated resultant in x direction = 1025.95V/m (this is equal to the E on P by A as E on P by B does not contribute to x direction.)

and

resultant in y direction = 2497.22(E on P by B in y direction) - 1521.03 (E on P by A in y direction)

Then I found magnitude (1416.17V/m) by pythag and angle using trig (44deg)
Heres a diagram, I know its messy

http://img402.imageshack.us/img402/83/img0351afn1.jpg [Broken]

Last edited by a moderator: May 3, 2017
7. Aug 10, 2007

### lightgrav

okay, except that phi is supposed to be from the +x -axis.

the E-field is caused by Charges, not points,
and the E doesn't "act on" a point (only acts on a charge).

8. Aug 10, 2007

### t_n_p

so phi = 180+44 = 224deg?

9. Aug 10, 2007

### learningphysics

I'm getting different numbers. For the x-component due to the +1 n C... I get

$$\frac{kq}{r^2} = \frac{(9*10^9)(1*10^{-9})}{52}$$

So that comes out to 9/52 = 0.173 V/m

then take the x - component... $$\frac{4}{\sqrt{52}}*0.173$$ which comes out to 0.096 V/m.

And for the y - component $$\frac{6}{\sqrt{52}}*0.173$$ which comes out to 0.144 V/m

I'm just giving the magnitudes, I'm ignoring the signs here...

10. Aug 10, 2007

### t_n_p

I was partially wrong, but I think you are too. (?)

Units are in cm, so root(52)=0.07cm
0.07^2 = 0.052

But im confused. Do you convert to m after squaring the square root of (52) or before?

11. Aug 10, 2007

### learningphysics

I made a mistake earlier. You should convert before squaring. You meant to write 0.07m right?

Last edited: Aug 10, 2007
12. Aug 10, 2007

### t_n_p

yes lol.

root (52) = 7.211cm
7.211^2 = 52cm
52cm = 0.52m
=17.29V/m

But the other calculation is right for Epb
((8.99*10^9)*(1*10^-9))/(0.06^2) = 2497.22V/m.

One is so big, one is so small, could this be right?

Last edited: Aug 10, 2007
13. Aug 10, 2007

### lightgrav

distances are in centi-meters ... not meters, so
9e9 [Vm/C]*1e-9[C]/.0052[m^2] = 1731 V/m ,
. . . yielding an x-component of 960 V/m ... hmm. roundoff error?

14. Aug 10, 2007

### t_n_p

How did you manage to get 0.0052?
(sqrt(52))²=52cm=0.052m?

15. Aug 10, 2007

### learningphysics

Sorry... I made a mistake in my post earlier that I edited afterwards. You should definitely convert to m before squaring.

16. Aug 10, 2007

### t_n_p

Sweet, thanks! Finally got that all sorted..

The x component then becomes 966.76 V/m
and y component then becomes -1433.28V/m

Resultant thus becomes
x direction = 966.76 V/m
y direction = 2497.22-1433.28 = 1063.94

and magnitude becomes 1437.57V/m.

My phi now changes to 180+48=228deg measured from positive x axis

17. Aug 10, 2007

### learningphysics

Cool! remember to convert your units before you do any squaring or anything... sorry about my earlier blunder.

18. Aug 10, 2007

### t_n_p

Not to worry! Thanks alot for helping in this thread and my other one! *bows to physics god*