PHYSICS HELP PLEASE!

Question

A large electroscope is made with "leaves" that are 78-cm-long wires with tiny 47 g spheres at the ends. When charged, nearly all the charge resides on the spheres.

If the wires each make a 26° angle with the vertical, what total charge Q must have been applied to the electroscope? Ignore the mass of the wires.

Answer 1

To work this problem, a picture would really help. The wires of the electroscope form an upside down V about a pivot point. The wires are connected to a pivot at the upper pinnacle and the ends of the wires closest to the ground have little spheres on their ends.

 

                                                           /\

                                                          /  \

                                                         /     \

                                                        /       \

                                                       O        O

 

If the spheres were not charged, the wires would just hang down. However, if the spheres each contain a charge Q of the same sign, there will be a repulsive force along a horizontal line between the spheres. This will overcome gravity and serve to push the spheres apart. Gravity will continue to push each sphere downward. Effectively, there will be a balance of torques in the position of the wires as described in the problem.

 

For a torque, one must calculate tau = r X F = rFsin(theta) where theta is the angle between the moment arm, which is the wire, and the force. Imagine dropping a vertical downward from the upper pivot point. Then you form two right triangles. We already know that the upper angle is 26 deg. Since the vertical forms a 90 deg angle with a line connecting the spheres, then the angle between the line connecting the spheres and the wire, or hypotenuse, is 90 - 26 = 64 deg. This is the angle between moment arm and the electric force in each wire.

 

What is the repulsive electric force? It is given by Coulomb's law: F = k(Q^2)/d^2 where d is the distance between the spheres. If the wires are 78 cm long, then

 

cos(64 deg) = (d/2)/78

 

d = 156 cos(64 deg) = 68.39 cm = 0.684 m

 

The Coulomb force F is

 

F = (9.0E9)(Q^2)/(0.684m)^2

 

F = (1.92E10)Q^2

 

The torque exerted by the electric force is tau_E = rXF = rFsin(64 deg) = (0.78 m)(1.92E10 Q^2)sin(64 deg)

tau_E = 1.35E10 Q^2 N-m

 

This electric torque must be balanced by the torque exerted by the weight of the sphere. The weight is straight down, or at an angle of 64 deg + 90 deg = 154 deg from the moment arm, or wire. The torque exerted by gravity is

tau_g = r X (mg) = rmgsin(154 deg) = (0.78 m)(0.047 kg)(9.8 m/s^2) sin(154 deg) = 0.157 N-m

 

These torques must balance.

 

tau_E = tau_g

 

1.35E10 Q^2 = 0.157

 

Q = 3.4E-6 Coulombs = 3.4 micro Coulombs

Answer 2

The spheres are being forced down by the force of gravity, yet apart by the charges of the two positively (or negatively) charged spheres.

 

So the distance from where the spheres would reside normally if there wasn't any sort of force on them is found by using some basic trig:

 

tan 27^o = (distance from equilibrium) / 78 cm

 

so our distance from equilibrium is

 

78 * tan 27^o = 39.74 and some change

 

Double that distance to get the distance between the two charged spheres:

 

39.74 cm * 2 = 79.49 cm

 

Our force between the two charged spheres are:

 

F_x = k* q₁ * q₂ / r² 

 

Keep in mind, q₁ and q₂ are identical in this instance, so I will be referring to both multiplied together as q²

 

F_x = 9 * 10^9 * q² / (0.7949)²

 

with that in mind, we now have to go back to the other half of this problem, the vertical component of the force.

 

F_y = mg

 

F_y = 0.047g * 9.8 m/s²

 

F_y = 0.4508 N

 

So to keep the force of gravity on the sphere equal to the force of the charge, you need to compare the x and y components of the force to the tension T on the wire.

 

The summation of all of our forces could be represented by:

 

T + F_x + F_y = 0 N (the spheres aren't moving, you see)

 

In addition, from this we know

 

T_x = -F_x

 

and

 

T_y = -F_y

 

(the signage isn't really important here, it's just to know that they're opposite of one another. We can reassign our frame of reference later. As such, I will be dropping the negative symbols, as they overcomplicate the remainder of this problem)

 

F_x = T * sin(27^o)          (NOT cosine, as the 26 degrees is from the VERTICAL, not HORIZONTAL. It helps to draw a picture)

 

F_y = T * cos(27^o)

 

solving F_y for T and making a substitution gives us

 

F_x = F_y * sin(27^o)/cos(27^o)

 

or

 

F_x = F_y * tan(27^o)

 

We now replace our values for F_x and F_y and solve for q.

 

F_x = 9 * 10^9 * q² / (0.7949)²

 

F_y = 0.4508 N

 

9 * 10^9 * q² / (0.7949)² = 0.4508 N * tan(27^o)

 

After much legwork on paper, you get the overall result of:

 

q = 4.016 * 10^-6 Coulombs

 

or

 

q = 4 microcoulombs (μC)