Answer:
Explanation:
The application of Gauss's law is used in the derivation as shown with detailed step by step in the attached file.
The potential difference on this spherical capacitor is ΔV = Va - Vb = kQ/a - kQ/b = kQ(1/a - 1/b)
We want to find how much momentum the dumbbell has at the moment it strikes the floor. Let's use this kinematics equation:
Vf² = Vi² + 2ad
Vf is the final velocity of the dumbbell, Vi is its initial velocity, a is its acceleration, and d is the height of its fall.
Given values:
Vi = 0m/s (dumbbell starts falling from rest)
a = 10m/s² (we'll treat downward motion as positive, this doesn't affect the result as long as we keep this in mind)
d = 80×10⁻²m
Plug in the values and solve for Vf:
Vf² = 2(10)(80×10⁻²)
Vf = ±4m/s
Reject the negative root.
Vf = 4m/s
The momentum of the dumbbell is given by:
p = mv
p is its momentum, m is its mass, and v is its velocity.
Given values:
m = 10kg
v = 4m/s (from previous calculation)
Plug in the values and solve for p:
p = 10(4)
p = 40kg×m/s
The magnetic force (Lorentz force) experienced by the proton in the magnetic field is given by

since

, because the velocity v and the force F in this problem are perpendicular, and so also the angle

between the velocity and the magnetic field B should be

.
Let's find the magnitude of the magnetic field; this is given by

To understand the direction, let's use the right-hand rule:
-index finger: velocity
- middle finger: magnetic field
- thumb: force
Since the velocity (index) points east and the force (thumb) points south, then the magnetic field (middle finger) points downwards. So we write:
B = -0.091 T