Modern family: Addition

State level Olympiad exam (4)

Q 3. A nonlinear circuit element can be made out of a parallel plate capacitor and small balls,
each of mass m, that can move between the plates. The balls collide inelastically with the
plates, dissipate all kinetic energy as thermal energy, and immediately release the charge they are carrying to the plate. Almost instantaneously, the balls then pick up a small charge of
magnitude q from the plate; the balls are then repelled directly toward the other plate under
electrostatic forces only. Another collision happens, kinetic energy is dissipated, the balls give
up the charge, collect a new charge, and the cycle repeats. There are n_0 balls per unit surface area of the plate. The capacitor has a capacitance C. The separation d between the plates is
much larger than the radius r of the balls. A battery is connected to the plates in order to maintain a constant potential difference V. Neglect edge effects and assume that magnetic forces and gravitational forces may be ignored.

(5) a. Determine the time it takes for one ball to travel between the plates in terms of any or all of the following variables: m, q, d, and V.

(5) b. Calculate the kinetic energy dissipated as thermal energy when one ball collides
inelastically with a plate surface in terms of any or all of the following variables: m,
q, d, and V.

(5) c. Derive an expression for the current between the plates in terms of the permittivity of
free space, ε_0 , and any or all of the following variables: m, q, n_0, C, and V.

(5) d. Derive an expression for the effective resistance of the device in terms of ε_0 , and any
or all of the following variables: m, q, n_0, C, and V.

(5) e. Calculate the rate at which the kinetic energy of the balls is converted into thermal
energy in terms of ε_0 , and any or all of the following variables: m, q, n_0, C, and V.

Solution 3 :

a. The electric field between the plates is given by E = V /d. The force on the charged ball is
then F = Eq = V q/d. The acceleration of the ball is a = (Vq)/(md).
Kinematics gives us d = (at^2)/2 for the time of flight.
So t = √(2d/a) = √{(2md^2)/(qV)}. (Q 3-1)

b. The kinetic energy collected by a ball will be K = qV as it moves between the plates. That’s
what will be dissipated.

c. The current is given by
I = ∆Q/∆t.
The total number of balls is N = n_0A, where A is the
surface area of a plate. The charge ∆Q is then ∆Q = n_0qA, so the current is
I = ∆Q/∆t
= n_0qA/ √{(2md^2)/(qV)}.  (Q 3-2)
We can’t stop here, since this is not in terms of the allowed variables. The problem is A and
d, but since C = €_0A/d, we have
I = n_0qA/√{(2md^2)/(qV)} ,
(Q 3-3)

= (A/d) (n_0q) √{(qV)/(2m)} ,
(Q 3-4)

= (C/€_0) (n_0q) √{(qV)/(2m)}.
(Q 3-5)

d. R = V /I,
so R = V/I = {(€_0V)/(Cn_0q)} √{(2m)/(qV)} . (Q 3-6)

We can simplify, slightly, with
R = {(€_0)/(Cn_0q)} √{(2mV)/q} . (Q 3-7)

e. P = V I, so
P = V {(C €_0)/(n_0q)}√{(qV)/(2m)}
= √{(€_0^2 n_0^2 C^2 q^3 V^3)/ (2m)} .

Q 4. A model of the magnetic properties of materials is based upon small magnetic moments
generated by each atom in the material. One source of this magnetic moment is the magnetic field generated by the electron in its orbit around the nucleus. For simplicity, we will assume
that each atom consists of a single electron of charge –e and mass m_e, a single proton of charge
+e and mass m_p >> m_e , and that the electron orbits in a circular orbit of radius R about the proton.

a. Magnetic Moments.

Assume that the electron orbits in the x-y plane.

i. Calculate the net electrostatic force on the electron from the proton. Express your
answer in terms of any or all of the following parameters: e, m_e, m_p, R, and the
permittivity of free space, ε_0 , where
ε_0 = 1/ (4kπ) .
(k is the Coulomb’s Law constant).

ii. Determine the angular velocity ω_0 of the electron around the proton in terms of
any or all of the following parameters: e, R, m_e and ε_0 .

iii. Derive an expression for the magnitude of the magnetic field B_e due to the orbital
motion of the electron at a distance z >>R from the x-y plane along the axis of
orbital rotation of the electron. Express your answer in terms of any or all of the
following parameters: e, R, z, R, m_e, ω_0 and the permeability of free space μ_0 .

iv. A small bar magnet has a magnetic field far from the magnet given by
B = (μ_0/2π) (m/z^3)
where z is the distance from the magnet on the axis connecting the north and south poles, m is the magnetic dipole moment, and μ_0 is the permeability of free
space. Assuming that an electron orbiting a proton acts like a small bar magnet,
find the dipole moment m for an electron orbiting an atom in terms of any or all of
the following parameters: e, me R, and ω_0 .

Solution 4a. Magnetic Moments

i. From Coulomb’s Law,
F = e^2/(4π€_0 R^2) (Q 4a-1)

ii. For circular motion,
F = (m_ev^2)/R
  = m_e R ω_0^2 ,  (Q 4a-2)

The force is provided by the Coulomb force, so
m_e R ω_0 ^2 = e^2/(4π€_0 R^2)
, (Q 4a-3)

ω0 = √{(e^2)/(4π €_0 m_e R^3)}
(Q 4a-4)

iii. From the law of Biot and Savart,
~ B_e = {(µ_0 i)/(4π)} (closed integral) {(~ds × ~r)/(r^3)} ,
(Q 4a-5)

B_e = {(µ_0i)/(4π)} (2πR) { R/(z^2 + r^2)^ 3/2} , (Q 4a-6)
≈ (µ_0 i R^2)/(2z^3) . (Q 4a-7)

For the current, i, we can write
i = q/t = eω_0/2π . (Q 4a-8)

Then
B_e = (µ_0eω_0R^2)/(4πz^3) .

iv. By substitution,
m = eω_0R/2 .