Magnitude and direction of force (electricity)?

Question

Hi,

Consider the three charges in Figure P25.42 http://i137.photobucket.com/albums/q208/infinitbelt/p25-42.gif in which d = 4.8 cm, q = 8 nC, and the positive x-axis points to the right. What is the force on the 5 nC charge? Give your answer as a magnitude and a direction.

? N
? degrees counterclockwise from the positive x-axis


I tried doing F = (kq1q2)/r^2, but I keep getting 2.94 x 10^-4 N, but that is wrong. Any ideas?


Thanks

Answer 1

You have to deal with a sum of forces.
The two positive charges will repel each other but attract the negative charge.
The force of their repulsion is irrelevant to this problem. Attraction of the negative charge is the main issue.

The force will be
Ft= F1 + F2 (vectors)
Q1= 8nC
Q2= -5nC
Q3= 5nC (The charge in question)
a=3cm, d=4.8cm
F1= kQ1Q3/(a^2+d^2) angle 180+arcTan(a/d)(negative x and y )
F2=kQ2Q3/(a^2) at 0 degreer or towards positive x axis

The total force is a sum of
Fx1=F1 sin (arcTan(a/d)
Fy1=F1 cos (arcTan(a/d)
Ftx=-Fx1 + F2 (the x component)
Fty= Fy1 (is the y component)

The angle in question is =(arcTan(Ffty/Ftx) "degrees counterclockwise from the positive x-axis"

Just do the math

Answer 2

You have to use that equation for the q-5nC and the -5nC-5nC pairs, and do the equation for both x and y directions. The -5 nC, 5 nC pair will simply be a force in the positive x-direction, but the q, 5nC pair will involve a component in the negative x direction and the negative y direction. You then have to add all the components and add your vectors to get the final answer