Electrical Forces, Torque & Flux

[kgigs6]

Homework Statement

A long, thin rod (length = 4.0 m) lies along the x axis, with its midpoint at the origin. In a vacuum, a +6.0 C point charge is fixed to one end of the rod, and a -6.0 C point charge is fixed to the other end. Everywhere in the x, y plane there is a constant external electric field (magnitude 5.0 x 103 N/C) that is perpendicular to the rod. With respect to the z axis, find the magnitude of the net torque applied to the rod.

Homework Equations

F=Eq
Torque = Fl

The Attempt at a Solution

F1 = (5E3N/C)(+6.0E-6C)
F1= 0.03N

F2= (5E3N/C)(-6.0E-6C)
F2= -0.03N

T1= (0.03N)(4.0m)
T1= 0.12 N/m

T2= (-0.03N)(4.0m)
T2= -0.12N/m

I'm not sure where to go from here in order to find the net torque, I don't think it is as simple as they cancel out but I can't figure out what to do next.

Homework Statement

A charge Q is located inside a rectangular box. The electric flux through each of the six surfaces of the box is: 1 = +4400 Nm2/C, 2 = +3530 Nm2/C, 3 = +1310 Nm2/C, 4 = -4520 Nm2/C, 5 = -5930 Nm2/C, and 6 = -4580 Nm2/C. What is Q?

Homework Equations

I'm not even sure where to start with this one, maybe one of these two equations:
Flux = Summation(Ecosphi)A
Flux = Q/Permittivity of free space

The Attempt at a Solution

 

[LowlyPion]

When you have 2 equal and opposite forces then I think the Torque on the object can be taken about one of the points of the application of the forces, and the torque would be the moment of the other force about that point. If your force is correct, (I didn't do the math), then either of your answers is correct, except that they should both end with the same direction, clockwise or counterclockwise and not different as you apparently show.

 

[nlightner]

I'm not sure where to start with this problem either. However, I think the first equation you listed may be more relevant in this case. Since the electric flux through each surface is given, we can use the formula Flux = EAcos(theta) where E is the magnitude of the electric field, A is the area of the surface, and theta is the angle between the electric field and the surface normal. We can then solve for Q by rearranging the equation to Q = Flux x Permittivity of free space. We would need to calculate the electric field for each surface using the given flux values and then use the fact that the electric field is constant inside the box to find the electric field at the location of the charge Q. From there, we can plug in the known values and solve for Q.