A small marble of charge q and mass m can slide without friction along a long, thin vertical rod passing through the center of a horizontal conducting ring of radius R, mounted on an insulating support. What is the magnitude of the minimum charge Q placed on the ring that would allow the marble to oscillate along the rod?

Solution:

- Concepts:

Electric field on the axis of a ring, stable equilibrium - Reasoning:

The force on Q is due to the field produced by Q and gravity. -
Details of the calculation:

Symmetry dictates that on the y axis only the y-component of the electric field is nonzero.

For a stable equilibrium we need e need F_{y}|_{equ}= 0, dF_{y}/dy|_{equ}≤ 0.

For the minimum charge Q we have dF_{y}/dy|_{equ}= 0.

F_{y}= kQqy/(y^{2}+ R^{2})^{3/2}- mg, k = 1/(4πε_{0}).

Equilibrium: F_{y }= 0 --> kQqy/(y^{2}+ R^{2})^{3/2}= mg,

Q = (mg/(kq))(y^{2}+ R^{2})^{3/2}/y, for any y ≠ 0 there is a solution for Q.

Stable equilibrium: dF_{y}/dy = -3kQqy^{2}/(y^{2}+ R^{2})^{5/2}+ kQq/(y^{2}+ R^{2})^{3/2}|_{equ}≤ 0.

3y^{2}> (y^{2}+ R^{2}), R^{2}< 2y^{2}, y > ±R/√2.

For the minimum charge we have Q = ±(mg2πε_{0}√(27) R^{2}/q.

y = +R/√2: Q = (mg2πε_{0}√(27) R^{2}/q, Q an q have the same sign.

y = -R/√2: Q = -(mg2πε_{0}√(27) R^{2}/q, Q an q have opposite signs.

An electron at a distance d = 1 mm is projected parallel to a grounded
perfectly conducting sheet with an energy of 100 electron volts. Let the
grounded conducting sheet lie in the xy-plane and the initial velocity of the
electron point in the x-direction. Let z(0) = d.

(a) Find the distance that the electron travels until it hits the plate.
Neglect the force of gravity.

(b) Find the magnitude and direction of a magnetic field parallel to the
surface of the plate and perpendicular to the electron velocity that keeps the
electron from hitting the plate.

Give numerical answers.

Hint: z = dsin^{2}θ may be a useful change of variable. ∫sin^{2}θdθ
= θ/2 - sin(2θ)/4.

Solution:

- Concepts:

The method of images - Reasoning:

We calculate the electrostatic force on the electron due to its image charge. - Details of the calculation:

(a) The force on the electron when it is a distance z from the plane is F_{z}= -kq_{e}^{2}/(2z)^{2}.

a_{x}= 0, v_{x }= constant = v_{0}= (2mE)^{½}= 5.93*10^{6}m/s.

md^{2}z/dt^{2}= mdv_{z}/dt = -kq_{e}^{2}/(2z)^{2}. mv_{z}dv_{z}/dt = -¼kq_{e}^{2}(1/z^{2})(dz/dt).

d/dt(½mv_{z}^{2}) = ¼kq_{e}^{2}d/dt(1/z).

½mv_{z}^{2}= ¼kq_{e}^{2}(1/z - 1/d).

(We can also use the work-kinetic energy theorem to obtain the same result.)

v_{z}= dz/dt = -[(kq_{e}^{2}/(2m)) (1/z - 1/d)]^{½}.

t_{ground}= ∫_{0}^{d}dz/[(kq_{e}^{2}/(2m)) (1/z - 1/d)]^{½}= (2md/kq_{e}^{2})^{½}∫_{0}^{d}dz(z/(d - z))^{½}.

Let z = dsin^{2}θ, dz = 2dsinθcosθdθ, z^{½}= d^{½}sinθ, (d - z)^{½}= d^{½}cosθ.

∫_{0}^{d}dz(z/(d - z))^{½}= 2d∫_{0}^{π/2}sin^{2}θdθ = 2dπ/4.

t_{ground}= (π/q_{e})(md^{3}/(2k))^{½}= 4.4*10^{-6}s.

During this time the electron travels 26.2 m in the x-direction.

(b) We need a magnetic field**B**= -B**j**pointing in the negative y-direction.

Then -q_{e}v**i**×(-B)**j**= q_{e}vB**k**.

q_{e}vB = kq_{e}^{2}/(2d)^{2}, B = kq_{e}/(4vd^{2}) = 6*10^{-11}T.

Three small metal charged balls have equal charges q and masses m, 4m and m.

The balls are connected by light non-conducting strings of length d each and
placed on a horizontal non-conducting frictionless table. Initially, the balls
are at rest and form a straight line as shown. Then, a quick horizontal
push gives the central ball a speed v directed perpendicular to the strings
connecting the balls.

(a) What is the total energy of the system?

(a) What is the kinetic energy associated with the motion of the CM?

(c) What is the minimum subsequent distance between the balls of mass m?

Solution:

- Concepts:

Conservation of energy and momentum. - Reasoning:

The kinetic energy of the CM of the system of interacting charges is constant. The kinetic energy about the CM changes when the potential energy changes. -
Details of the calculation:

(a) We treat the push as an impulse. Right after the push we have for the system T = ½4mv^{2}, p = 4mv, U = 2kq^{2}/d + kq^{2}/(2d).

E_{total}= 2mv^{2}+ 2kq^{2}/d + kq^{2}/(2d).

(b) T_{CM}= p^{2}/(2m_{total}) = 16mv^{2}/(12m) = 4mv^{2}/3.

(c) When the system has no kinetic energy about the CM, the potential energy is highest and the distance between the balls of mass m is smallest.

2mv^{2}+ 2kq^{2}/d + kq^{2}/(2d) = 4mv^{2}/3 + 2kq^{2}/d + kq^{2}/d_{min}.

2mv^{2}/3 = kq^{2}(1/d_{min}- 1/(2d)),

d_{min}= 2d/[1 + 4mv^{2}d/(3kq^{2})] = 2d/[1 + 16πε_{0}mv^{2}d/(3q^{2})].

A capacitor composed of two parallel infinite conducting sheets separated by
a distance d is connected to a battery. The lower plate is maintained at some
potential V_{1} and the upper plate is maintained at some potential V.
A small hemispherical boss of radius a << d is introduced on the lower plate.
State the boundary conditions for this problem. (Hint: Consider the limit as
the distance between the plates becomes large.) Find the potential between the
plates and the surface charge density on the plates.

Solution:

- Concepts:

Boundary value problems, azimuthal symmetry. - Reasoning:

Put the origin of the coordinate system at the center of a sphere, the upper half of which is the hemispherical boss. Let the z-axis point up.

Then Φ(r,θ) = ∑_{n=0}^{∞}[A_{n}r^{n}+ B_{n}/r^{n+1}]P_{n}(cosθ) is the most general solution in the region between the plates, since the charge density ρ = 0 there. To find the specific solution we apply boundary conditions. - Details of the calculation:

The boundary conditions are Φ = V_{1}on the lower plate and the boss and Φ = V on the upper plate.

Φ = V_{1}+ A_{1}'rcosθ + (B_{1}'/r^{2})cosθ between the plates outside the boss.

We assume all other coefficients are zero. If we find a solution with this assumption, it is the only solution.

Boundary conditions:

(i) Φ is continuous at r = a. 0 = A_{1}'a + B_{1}'/a^{2}. B_{1}' = -A_{1}'a^{3}.

(ii)**E**= -**∇**Φ =**E**_{0}at r >> a, i.e. far away from the hemispherical boss.

-A_{1}'cosθ = E_{0}cosθ, A_{1}'sinθ = -E_{0}sinθ, A_{1}' = -E_{0}, B_{1}' = E_{0}a^{3}. E_{0}= (V - V_{1})/d.

Therefore

Φ = V_{1}- E_{0}rcosθ + (E_{0}a^{3}/r^{2})cosθ.

E_{r}= E_{0}cosθ + 2(E_{0}a^{3}/r^{3})cosθ, E_{θ}= -E_{0}sinθ + (E_{0}a^{3}/r^{3})sinθ.

**E**=**E**_{0}+ 2(E_{0}a^{3}/r^{3})cosθ (**r**/r) + (E_{0}a^{3}/r^{3})sinθ (**θ**/θ).

**E**= external field + field due to polarized sphere.

Surface charge on the plates: σ = ε_{0}**E·n**.

**E·n**= -E_{0}on the upper plate, σ = -ε_{0}E_{0}.

**E·n**= -E_{θ}= E_{0}- (E_{0}a^{3}/r^{3}) on the flat part of the lower plate, σ = ε_{0}E_{0}- ε_{0}(E_{0}a^{3}/r^{3}).

**E·n**= E_{r}= 3E_{0}cosθ on the boss, σ = ε_{0}3E_{0}cosθ.

An aperture of radius a in a thin plate separates a region in which the electric
field is E_{1} from a region in which it is E_{2}. E_{1}
and E_{2} are perpendicular to the plate. Let E_{1} > E_{2}. A beam of particles of charge q
and energy qV_{0} comes to a focus at a distance z_{1} in
front of the plate, goes through the aperture, and comes to a second focus at a distance
z_{2}
behind the aperture.

Show that 1/z_{1} + 1/z_{2} = (E_{1}
- E_{2})/4V_{0}.

Assume
V_{0} >> E_{1}z_{1}, E_{2}z_{2}; a
<< z_{1}, z_{2}.

Hint: Use Gauss' law to estimate the radial impulse given to a particle by the field near
the aperture.

Solution:

- Concepts:

Gauss' law, impulse =**F**_{avg}Δt. - Reasoning:

Gauss' law can be used to find the average radial component of the field near the aperture. This radial component results in a radial component of the force, which delivers the radial impulse. - Details of the calculation:

Assume the field is uniform, except in a small region Δz near the aperture. Consider a Gaussian surface a shown.

E_{1}πr^{2}- E_{2}πr^{2}+ E_{r_avg}2πrΔz = 0. (Gauss' law)

E_{r_avg}= -½r(E_{1}- E_{2})/Δz. Assume E_{1}> E_{2}.

We have cylindrical symmetry.

Let a particle have an initial velocity**v**= v_{z}**k**+ v_{y}**j**at z = -z_{1}. Let v_{z}>> v_{y}.

Since V_{0}>> E_{1}z_{1}, E_{2}z_{2}we can assume that v_{z}≈ constant.

The particle reaches the aperture in time t = z_{1}/v_{z}.

Its y-coordinate at the aperture is y = v_{y}t = v_{y}z_{1}/v_{z}.

It receives an impulse**F**_{avg}Δt = -½qy(E_{1}- E_{2})Δt/Δz**j**= Δp_{y}**j**.

Δv_{y}= Δp_{y}/m = -qy(E_{1}- E_{2})/(2mv_{z}), since Δz = v_{z}Δt.

Δv_{y}= -qv_{y}z_{1}(E_{1}- E_{2})/(2mv_{z}^{2}).

The particles new y-component of the velocity is v_{y}+ Δv_{y}.

In time t' it reaches the z-axis. 0 = y + v_{y}t' + Δv_{y}t'.

t' = -y/(v_{y}+ Δv_{y}) = -(z_{1}/v_{z})2mv_{z}^{2}/(2mv_{z}^{2}- qz_{1}(E_{1}- E_{2})).

z_{2}= v_{z}t' = -2z_{1}mv_{z}^{2}/(2mv_{z}^{2}- qz_{1}(E_{1}- E_{2})) = -4z_{1}qV_{0}/(4qV_{0}- qz_{1}(E_{1}- E_{2})),

since qV_{0}≈ ½mv_{z}^{2}.

1/z_{2}= -(4qV_{0}- qz_{1}(E_{1}- E_{2}))/4z_{1}qV_{0}= -1/z_{1}+ (E_{1}- E_{2}))/(4qV).

1/z_{1}+ 1/z_{2}= (E_{1}- E_{2})/4V_{0}.