Problem 1:
A circuit below consists of four resistors connected by ideal wires. The
circuit draws power P if an ideal battery is connected to either points A and D
or points B and C. If the same battery is connected to either points A and B or
points C and D, the circuit draws power 2P. What power would be drawn if the
same battery is connected to points A and C?
Solution:
- Concepts:
Resistors in series and parallel
- Reasoning:
The power drawn by a circuit is P = V2/Requ.
We have 4 equations for Requ that can be solved for the
resistance of the 4 resistors in terms of the battery voltage V and P.
- Details of the calculation:
Label the resistors as shown in the figure below.
Battery connected to points A and D:
1/Requ = (R1 + R2 + R3 + R4)/(R4(R1
+ R2 + R3)) = Rtotal/(R4(R1
+ R2 + R3)) .
Battery connected to points B and C: 1/Requ = Rtotal/(R2(R1
+ R4 + R3)) .
same power --> R4 = R2.
Battery connected to points A and B: 1/Requ = Rtotal/(R1(R2
+ R3 + R4)).
Battery connected to points C and D: 1/Requ = Rtotal/(R3(R2
+ R1 + R4)).
same power --> R1 = R3.
2P = V2Rtotal/R1(R2 + R1
+ R2)), P = V2Rtotal/R2(R1
+ R2 + R1)).
ratio: 2 = R2(R2 + 2R1)/(R1(R1
+ 2R2)).
R2 = (1 + √3)R1.
P = V2(2 + √3)/(R1(3 + 2√3)). V2/R1
= P (3 + 2√3)/(2 + √3).
Battery connected to A an C:
1/Requ = 2/(R1 + R2)) = 2/(R1(2
+ √3)).
PAC = 2V2/(R1(2 + √3)) = P 2(3 + 2√3)/(2 + √3)2
= 0.928 P.
Problem 2:
Two conducting spheres of radii a and b < a, respectively, are connected by a
thin metal wire of negligible capacitance. The centers of the two spheres are
at a distance d >> a > b from each other.
A total net charge Q is located on the system. Evaluate to zeroth order
approximation, neglecting the induction effects on the surfaces of the two
spheres,
(a) how the charge Q is partitioned between the two spheres,
(b) the value V of the electrostatic potential of the system
(assuming zero potential at infinity) and the capacitance C = Q/V,
(c) the electric field at the surface of each sphere,
comparing the magnitudes and discussing the limit b --> 0.
Solution:
- Concepts:
Conductors in electrostatics
- Reasoning:
The surface of a conductor is an equipotential surface.
- Details of the calculation:
(a) To the zeroth order in a/d and b/d, we assume the surface charges to be
uniformly distributed.
The electrostatic potential generated by each sphere outside its volume is thus
equal to the potential of a point charge located at the center of the sphere.
Let us denote by Qa and Qb the charges on each sphere,
with Qa + Qb = Q. The charge on the wire is negligible
because we have assumed that its capacitance is negligible. The electrostatic
potentials of the spheres with respect to infinity are
Va = keQa/a, Vb = keQb/b.
Va = Vb --> Qb = Qab/a, Qa(1
+ b/a) = Q, Qa = Qa/(a + b), Qb = Qb/(a + b).
ke = 1/(4πε0).
(b) V = keQ/(a + b), C = Q/V = (a + b)/ke.
(c) The electric field at the surface of each sphere is perpendicular to the
surface of that sphere. If Q is positive, it points radially away from the
center of the sphere.
For the magnitudes we have Ea = keQa/a2
= keQ/((a + b)a), Eb = keQb/b2
= keQ/((a + b)b).
The smaller the radius of curvature b, the stronger is the electric field on the
surface of the sphere with radius b. As b --> 0, we no longer have a sphere,
just the tip of the wire with a very strong electric field at the surface of the
tip.
Problem 3:
An electric dipole p is
located at a distance r from a point charge q, as shown in the figure.
The angle between p and r is θ. Evaluate the electrostatic force
on the dipole.
Solution:
- Concepts:
Newton's 3rd law, the dipole field, E(r)
= [1/(4πε0)](1/r3)[3(p∙r)r/r2 - p].
- Reasoning:
Let us calculate the force the dipole exerts on q.
By Newton's 3rd law F12 = -F21.
- Details of the calculation:
Choose your coordinate system as shown.
Then the dipole points in the
z'-direction and we can write
E = (p/(4πε0r3))[2cosθ'
er + sinθ' eθ'],
with θ' = π - θ.
The force on the charge q is
F =
qE.
Fon q = (qp/(4πε0r3))[2cosθ'
er + sinθ' eθ'],
The force on the dipole is
Fon p = (qp/(4πε0r3))[-2cosθ'
er - sinθ' eθ'],
or Fon p
= (qp/(4πε0r3))[2cosθ
er - sinθ eθ')],
In terms of the original vector
r and a unit vector k pointing up,
we have
Fon p = (qp/(4πε0r3))[-2cosθ
(r/r) + sinθ k)].
The dipole is pulled towards the charge and up.
A different approach:
The potential energy of an intrinsic magnetic dipole in a magnetic field is U =
-p∙E.
The force on the dipole on an external magnetic field is
F = -∇U
= ∇(p∙E) = (p∙∇)E,
since p is constant.
At t = 0 put q at the origin and the dipole on the x-axis at x = r and let the
z-axis point up.
The electric field produced by the charge in the x-z plane (for x > 0) is
E
= (q/(4π ε0r2))er.
Fon p
= (px∂E/∂x + pz∂E/∂z)|x=x0,
y=0, z=0.
While on the positive x-axis E points in the x-direction, we need to know
E in the neighborhood
of the x-axis to evaluate the derivatives.
E = (q/(4π ε0r2))
i x/r + k z/r) = (q/(4π ε0r3)) (-xi + zk) = A(r) (xi + zk).
∂E/∂x = (∂A/∂x)(xi + zk) - Ai.
∂E/∂z = (∂A/∂z)(xi + zk) + Ak.
∂A/∂x = (q/(4πε0))∂(x2 +
y2 + z2)-3/2/∂x) = -(3q/(4πε0r5))x.
∂A/∂z = (q/(4πε0))∂(x2 +
y2 + z2)-3/2/∂x) = -(3q/(4πε0r5))z.
∂E/∂x|x=r, y=0, z=0 = -(3q/(4πε0r5))r2
i + (q/(4πε0r3))
i = -(2q/(4πε0r3))
i.
∂E/∂z|x=r, y=0, z=0 = (q(4πε0r3))
i.
Fon p = (q/(4πε0r3))
[-2px i + pz k)] =
(qp/(4πε0r3))
[-2cosθ i + sinθ k)].
Problem 4:
The volume between two concentric spherical surfaces of
radii a and b (a < b) is filled with an inhomogeneous dielectric with
permittivity
ε = ε0/(1 + κr),
where ε0
and κ are constants and r is the radial coordinate.
Thus D(r) = εE(r).
A charge Q is placed on the inner surface, while the outer surface is grounded.
Find:
(a) The displacement D(r) and the field E(r) in the region a < r
< b.
(b) The capacitance of the device.
(c) The volume polarization charge density in the region a < r < b.
(d) The surface polarization charge density at r = a and at r = b.
Solution:
- Concepts:
Gauss law and symmetry
- Reasoning:
The problem has spherical symmetry. We can find the field between the surfaces
in terms of the free charge Q on the surface using Gauss' law for D.
- Details of the calculation:
(a) Gauss law for D: D(r) = Q/(4πr2)
er.
E(r) = Q(1 +
κr)/(4πε0r2) er.
(b) C = Q/ΔV.
ΔV = Va =
∫abE∙dr = [Q/(4πε0)](1/a
- 1/b + κ ln(b/a)).
C = (4πε0ab)/(b - a + abκ
ln(b/a)).
(c) D = ε0E + P
P = (ε - ε0)E = (ε - ε0) Q(1 + κr)/(4πε0r2)
er = -Qκ/(4πr) er.
The volume polarization charge density is
ρP = -∇∙P.
ρP = Qκ/(4πr2).
Total volume polarization charge = Qκ(b - a)
(d) σP = P∙n.
At r = a, n = er. σP = Qκ/(4πa).
At r = b, n = -er. σP = -Qκ/(4πb).
The total polarization charge is zero, as required.
Problem 5:
A rectangular block of dielectric material with permittivity ε is partially
inserted between two parallel plane conducting plates. The plates are
square, of side l, and are separated
by a distance d, with d << l. The
dielectric is also square, of side l,
and has a thickness of almost d. A potential V0 is applied
across the plates by a battery.
(a)
When the dielectric has a length x inserted between the plates, calculate
the force on the dielectric, including its direction.
If x is now changed, does this force depend on x?
(b)
When the dielectric has a length x inserted between the plates, the
battery is disconnected. Calculate
the force on the dielectric, including its direction.
If x is now changed, does this force depend on x?
Solution:
- Concepts:
Capacitance, energy and work
- Reasoning:
Fx = -dWmech/dx. We must find the work done by
external agents as a dielectric is inserted into a capacitor and the capacitance
changes.
- Details of the calculation:
(a) Capacitance as a function of x:
C = εlx/d + ε0l(l
-x)/d.
Total energy store in the capacitor: U = ½CV02.
dWmech + dWbat = dU. dWbat = dQV0
= dCV02. dU = ½dCV02.
dWmech = -½dCV02.
Fx =
-dWmech/dx = ½V02dC/dx = ½V02(ε
- ε0)l/d.
The dielectric is pulled towards the right, into the capacitor.
The force is independent of x.
(b) When the battery is
disconnected with the dielectric inserted a distance x, the force is
Fx
= ½V02(ε - ε0)l/d.
At that moment the voltage and the distances are the same as for case
(a).
But when x changes, Q, not V,
stays constant. We therefore write
Fx = ½(Q0/C)2(ε - ε0)l/d,
where Q0 = CinitialV0.
Fx = ½Q02(ε
- ε0)ld/(εlx + ε0l(l -x))2. Since C depends
on x, Fx depends on x.
Note:
We can also find Fx for part (b) by writing dWmech
= dU, U = ½Q02/C.
Fx = -dWmech/dx = -dU/dx =
½(Q02/C2)dC/dx
= ½(Q0/C)2(ε - ε0)l/d.