1. Jan 13, 2008

### BuBbLeS01

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

3. 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????

2. Jan 13, 2008

### 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.

3. Jan 13, 2008

### BuBbLeS01

Don't I need that to find F???

4. Jan 13, 2008

### 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.

5. Jan 13, 2008

### BuBbLeS01

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

6. Jan 13, 2008

### Shooting Star

Yes. Now find the couple.

7. Jan 13, 2008

### 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

8. Jan 13, 2008

### 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?

9. Jan 13, 2008

### BuBbLeS01

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

10. Jan 13, 2008

### 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.

11. Jan 13, 2008

### 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...

12. Jan 13, 2008

### 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.

13. Jan 13, 2008

### BuBbLeS01

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

14. Jan 13, 2008

### Shooting Star

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.)

15. Jan 13, 2008

### BuBbLeS01

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

16. Jan 13, 2008

### 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.