I can't figure out what I am doing for nothing Max Torque

[BuBbLeS01]

I can't figure out what I am doing for nothing! Max Torque...Please Help!

Homework Statement

Two small plastic spheres, one charged to 18.95 nC and the other to -18.95 nC, are connected by a 50-mm-long insulating rod. Suppose this dipole is placed a uniform electric field with strength 8.130×10^5 N/C. What are the maximum possible torque on the dipole?

Homework Equations

The Attempt at a Solution

r = 0.050m
q1 = 18.95 nC
q2 = -18.95 nC
E = 8.130 e^5

Torque = r*F(perpendicular)
F = K * q1 * q2 / r^2 = 0.00129N
Toqrue 1 = 0.00129 * (0.050/2) = 3.23 e^-5
Toqrue 2 = 0.00129 * (0.050/2) = 3.23 e^-5
T1 + T2 = 6.46 e^-5 N*m

Motion is clockwise around pivot in center...

Ahhh what am I doing wrong?

 

[Shooting Star]

> F = K * q1 * q2 / r^2 = 0.00129N

Why do you need this?

Torque in this case is a couple, which is equal to the one of the perpendicular forces into distance between the two points of application.

 

[BuBbLeS01]

Don't I need that to find F?

 

[Shooting Star]

By F if you mean the force between the two charges, then no. What you do have to find is the force on a charge by the field.

 

[BuBbLeS01]

So that would this equation?
F = q*E
F = 18.95nC * 8.130×10^5 N/C = 1.54 x 10^-2 N

 

[Shooting Star]

Yes. Now find the couple.

 

[BuBbLeS01]

T = rF(perp)
T1 = (0.050/2) * 1.54 x 10^-2 = 3.85 x 10^-4
T2 = (0.050/2) * 1.54 x 10^-2 = 3.85 x 10^-4
T1 + T2 = 7.7 x 10 ^-4 N*m

 

[Shooting Star]

Only one of T1 or T2 will do. You need not add them.

Your original Q was "What are the maximum possible torque on the dipole?"

When do you think is the torque maximum?

 

[BuBbLeS01]

When they are perpendicular? I am not sure how to answer that?

 

[Shooting Star]

That's right. It has not been asked, but you should know. In general, the torque will be T1*r*sin(theta), if theta is the angle between the rod and the field direction. So, when theta=90 deg, the torque is max.

 

[BuBbLeS01]

Oh ok, thanks. So the answer would just be 1 of the torques, 3.85 x 10^-4 N*m? Why don't we add the torques? I thought...
Tnet = T1 + T2 + T3...

 

[Shooting Star]

The net torque has to be the sum of all torques around one point. Since T2 passes through the point q2, the torque of T2 is zero about that point.

Try and take the sum of the torques about the mid point, and see what result you get.

 

[BuBbLeS01]

It says 3.85 x 10^-4 N*m is wrong?

 

[Shooting Star]

T = rF(perp)
T1 = (0.050/2) * 1.54 x 10^-2 = 3.85 x 10^-4
T2 = (0.050/2) * 1.54 x 10^-2 = 3.85 x 10^-4
T1 + T2 = 7.7 x 10 ^-4 N*m

You have taken the torque around the mid point, after all, and added, and so T1 + T2 should be the correct answer. (You have done 0.050/2, which I did not notice.)

Check for the power of 10.

(The answer is 7.7*10^-8 Nm? I may be wrong. Don't make me do arithmetic.)

 

[BuBbLeS01]

oh i thought i wasn't supposed to add them?

 

[Shooting Star]

You are always supposed to add all the torques to get the net torque. I had asked you to take the torque about one of the ends of the rod, so that you only have to find the torque of the other force about this point (refer post #2), because the other torque would be zero (refer post #12). I was under the impression that you knew what a couple was.

Now that you have taken the torques about the mid point (refer post #7), then you have to add them to get the net torque. Ir slipped me in post #7 that you had taken r/2, instead of r.

You have done correctly.