Discussion in 'Homework Help' started by smarch, Jul 3, 2010.

1. ### smarch Thread Starter Active Member

Mar 14, 2009
52
0
Can someone please help me with the question that is attached as a JPEG to this post.

I have worked out:

F3 = -11.24 x 10 ^-6
F1 = 16.86 x 10^-6

I am now unsure of how to carry on with the rest of the quesiton.
To find the magnitude of q2 do I use the cosine rule?

And to find the direction do I use :

a/sina = b/sinb

Any help would be appreciated, thanks.

File size:
39.6 KB
Views:
19
2. ### Georacer Moderator

Nov 25, 2009
5,150
1,271
Smarch, apparently you forgot to multiply the Electric Field with the test charge in order to find the force exerted on it. The values you found correspond to the Magnitude of the Electric field at the point (0.04,0).
Remember that the formula for the electrostatic force is:
$F=\frac{Q \cdot q}{4 \cdot \pi \cdot {\epsilon}_o \cdot r^2}$
Also notice that one force is repulsive and one is attractive

When you have found the forces, make a diagram.
The x component of the total force is $F_x = F_1 \cdot \cos(\phi_1)\ +\ F_2 \cdot \cos(\phi_2)$
and the y component is the same formula with but with sine.
In order to find the magnitude and angle of the total force use:
$|F|=sqrt{{F_x}^2\ +\ {F_y}^2}\ and\\
\angle F=\tan^{-1}(F_y\ -\ F_x)$

I guess that's all there is.