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Solutions to problems related to vector derivatives and electric fields from IIT Kanpur's PHY103A: Physics II course, Semester II, 2017-18. It includes the vector derivatives in cartesian, spherical, and cylindrical coordinates, vector identities, and calculations of electric fields and charges.
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PHY103A: Physics II Semester II, 2017-18; IIT Kanpur
Feb 22nd, 2018 Time: 8:00-10:00 am Maximum Marks: 80
Answer all 5 questions. Calculators are not allowed. A set of important formulas and information are provided on the next page.
The vector derivatives in cartesian coordinates:
∇T =
∂x xˆ +
∂y ˆy +
∂z ˆz
∇ · V = ∂Vx ∂x
∂Vy ∂y
∂Vz ∂z ∇ × V =
∂Vz ∂y − ∂Vy ∂z
ˆx +
∂Vx ∂z − ∂Vz ∂x
ˆy +
∂Vy ∂x − ∂Vx ∂y
ˆz
∂x^2 +^
∂y^2 +^
∂z^2
The vector derivatives in spherical coordinates:
∇T =
∂r rˆ^ +
r
∂θ θˆ + 1 r sin θ
∂φ φ,ˆ
∇ · V =
r^2
∂r
r^2 Vr
r sin θ
∂θ
sin θVθ
r sin θ
∂φ Vφ
∇ × V = 1 r sin θ
∂θ (sin θVφ) − ∂Vθ ∂φ
r ˆ +^1 r
sin θ
∂Vr ∂φ
∂r (rVφ)
θˆ +^1 r
∂r (rVθ ) − ∂Vr ∂θ
φˆ
∇^2 T =
r^2
∂r
r^2
∂r
r^2 sin θ
∂θ
sin θ
∂θ
r^2 sin^2 θ
∂φ^2
The vector derivatives in cylindrical coordinates:
∇T =
∂s sˆ +
s
∂φ φˆ + ∂T ∂z ˆz
∇ · V =
s
∂(sVs) ∂s
s
∂Vφ ∂φ
∂Vz ∂z ∇ × V =
s
∂Vz ∂φ
∂Vφ ∂z
s ˆ +
[ (^) ∂Vs ∂z
∂Vz ∂s
φˆ +^1 s
[ (^) ∂(sVφ) ∂s
∂Vs ∂φ
z ˆ
∇^2 T =^1 s
∂s
s ∂T ∂s
s^2
∂φ^2
∂z^2
Vector Identities:
∇ · (f A) = f (∇ · A) + A · (∇f ) ∇ · (A × B) = B · (∇ × A) − A · (∇ × B) ∇ × (f A) = f (∇ × A) − A × (∇f ), ∇ × (A × B) = (B · ∇)A − (A · ∇)B + A(∇ · B) − B(∇ · A) ∇ × (∇ × A) = ∇(∇ · A) − ∇^2 A.
Fundamental Constants:
0 = 8. 85 × 10 −^12 C^2 /Nm^2 μ 0 = 4π × 10 −^7 N/A^2
Problem 2: (a) A dielectric sphere of radius r, centered at the origin, carries a polarization P = kr. Calculate the total volume bound charge and the total surface bound charge. (3 marks)
(b) Consider a planar square sheet of side a with constant surface charge density σ (see figure below). Calculate the electric field on the z-axis very close to the sheet, that is, when s → 0. (3 marks)
y x
z
s
(c) A sphere of radius R, centered at the origin, carries charge density ρ(r, θ) = k
r^2 (R−r) sin θ, where k is a constant, and r, θ are the usual spherical coordinates (see figure below). Evaluate the leading contribution to the potential at points r 0 on the z axis, far from the sphere. (4 marks)
y
x
z
Solution 2: (a) The volume charge density is given by ρb = −∇ · P = −∇ · (kxˆx + kyˆy + kzˆz) = −k(1 + 1 + 1) = − 3 k. So the total volume charge is Qv = − 3 k
πR^3 = − 4 πkR^3. The surface charge density is given by σb = P · ˆn|surface = kR. So, the total surface charge is Qs = kR × 4 πR^2 = 4 πkR^3. So, the total bound charge is Q = Qv + Qs = − 4 πkR^3 + 4πkR^3 = 0, as expected.
(b) Since the observation point is very close to the surface, we have s → 0. Using the Gauss law and the boundary condition that Eabove − Ebelow = σ/ 0 , we have that the electric field on the z-axis very close to the sheet is σ 2 0 (c) Monopole term: We have
Vmono(r 0 ) =
4 π 0 z
ρ(r)dτ
=
4 π 0 z
k
r^2 (R − 2 r) sin θ
r^2 sin θdrdθdφ
4 π 0 z kR
(R − 2 r) sin^2 θdrdθdφ = 1 4 π 0 z kR ×
0
(R − r)dr ×
∫ (^) π
0
sin^2 θdθ
∫ (^2) π
0
×dφ
The r-integral yields
0 (R^ −^ r)dr^ = (Rr^ −^ r
R 0 = R^2 − R^2 /2 = R^2 /2. Therefore, the leading contribution is
Vmono(r 0 ) =
4 π 0 z kR ×
π 2 × 2 π = πkR^3 8 0 z
Problem 3: (a) Two long coaxial cylindrical metal tubes (inner radius a, outer radius b) stand vertically in a tank of dielectric oil (susceptibility χe, mass density ρ). The inner one is maintained at potential V and the outer one is grounded. To what height (h) does the oil rise in the space between the tubes. (10 marks)
(b) A cylindrical conductor of radius a is centered about the z axis. It has a cylindrical hole of radius b bored parallel to, and centered a distance d from, the z axis (d + b < a). The current density J = J zˆ is uniform throughout the remaining metal of the cylinder and is parallel to the axis. A cross-section of the cylinder is shown in the figure below. Find the magnitude and the direction of the magnetic field in the hole. (10 marks)
Solution 3: (a) We need to find the capacitance as a function of the height h and then use the formula for the force on a dielectric to obtain the final height that the oil would rise up to. In order to calculate the capacitance we’ll need to calculate the potential and then the total charge. Assume that the total length of the tubes is l and that the oil rises up to height h. We assume that the line charge density of the air-part of the tubes is λ and that of the oil part is λ′. The potential can be calculated as follows:
For the air part Eair = 2 λ 4 π 0 s ⇒ Vair = 2 λ 4 π 0 ln
b a
For the oil part Doil = 2 λ′ 4 πs ⇒ Eoil = 2 λ′ 4 πs ⇒ Voil = 2 λ′ 4 π ln
b a
We have the boundary condition that Voil = Vair = V. Therefore we have λ′^ =
0 λ^ =^ r^ λ. The total charge on the tube is Q = λ′h + λ(l − h) = r λh − λh + λl = λ[(r − 1) + l] = λ(χeh + l)
Therefore the capacitance of the system is
C =
λ(χeh + l) 2 λ 4 π 0 ln
b a
) = 2π 0 (χeh + l) ln
b a
Therefore, the force on the oil in the upward direction is
F 1 =
dC dh
2 π 0 χe ln
b a
Problem 4: (a) A uniform surface charge density σ exists over the entire x − y plane except for a circular hole of radius R centered at the origin. Find the electric field along the z-axis. Plot the electric field strength as a function of z in the range −∞ < z < ∞, indicating all the limiting values on the plot. (6+2 marks)
(b) The electric field due to a static charge distribution has x component Ex = kx^2 yz. Construct a valid functional form for the other two components. (6 marks) (c) An infinite conducting plane has a hemispherical bump of radius R (see figure below). A point charge q is located at a distance R above the top of the hemisphere. Find the force on the charge q. (6 marks)
z
y
Solution 4: (a) First, we calculate the field due to a uniformly charged disk centered at the origin. The field will be in the z-direction. And we have
q z
da
Ez =
4 π 0
σda r^2 + z^2
cos θ
=
4 π 0
σ r^2 + z^2
z √ r^2 + z^2
rdrdφ
= σz 4 π 0
0
rdr (r^2 + z^2 )^3 /^2
∫ (^2) π
0
dφ
σz 2 0
0
rdr (r^2 + z^2 )^3 /^2 = σz 2 0
z^2
z^2 + R^2
Thus we have for the field due to a uniformly charged disc is:
Edisc = σz 2 0
z
z^2 + R^2
z ˆ (8)
We know that the field due to a uniformly charged plane is
Eplane = σ 2 0 z ˆ (9)
Therefore, the field due to the plane with a circular hole of radius R
E = Eplane − Edisc = σ 2 0
z √ z^2 + R^2
zˆ (10)
Now, in the limit z → ±∞, we have E = ± σ 2 0
(b) We will solve this problem using the image charge method.
z
y
q
q ''' =-q
q '' =-q '
q ' d d
We have q′^ = image charge due to the hemisphere = − qR 2 R
q 2
Location of the image charge d =
We need to put two more charges due to the infinite plane. q′′^ = −q′^ = q 2 at z = −
q′′′^ = −q at z = − 2 R.
Therefore, the force Fq on q is
Problem 5: A spherical shell of radius a carries a surface charge density σ. The shell rotates around the z-axis at an angular velocity ω in an external magnetic field B = B 0 xˆ. (a) Find the surface current density on the shell. (6 marks)
(b) Calculate the magnetic dipole moment m of the rotating spherical shell. (6 marks)
(c) Calculate the torque N exerted on the shell using the basic definition of the torque (N = r × F ). (6 marks)
(d) Again, calculate the torque N exerted on the shell using the formula N = m × B. (2 marks)
Solution 5:
μ
z (^)!
y x Á
ad μ r
(a) The radius of the shell is a and the constant surface charge density is σ. When the spherical shell spins with an angular frequency ω the surface current density K is K = σV = σ(ω × r). (17) Here, r is the vector from the origin to the element on the shell. Therefore, we have
K = σ(ω × r) = σ
x ˆ yˆ zˆ 0 0 ω a sin θ cos φ a sin θ sin φ a cos θ
∣∣ =^ σωa^ sin^ θ(cos^ φ^ yˆ^ −^ sin^ φ^ xˆ) =^ σωa^ sin^ θ^ φˆ^ =^ K^0 sin^ θ^ φ,ˆ (18) where K 0 = σωa.
(b) The current on the infinitesimal element on the shell is I = K 0 sin θadθ. The perpendicular distance from the element to the z−axis is ρ = a sin θ. So, the infinitesimal dipole moment due to the ring is given by
dm = Iπρ^2 zˆ = πK 0 sin^3 θa^3 dθ zˆ. (19) Therefore the total dipole moment is
m =
dm =
∫ (^) π
0
πK 0 sin^3 θa^3 dθ zˆ =
πa^3 K 0 zˆ = 4 π 3 σa^4 ω zˆ (20)
(c) The force dF acting on a current element dl carrying current I in a magnetic field B = B 0 xˆ is given by dF = Idl × B = Kda × B = (K × B)da (21)
The torque on this element is dN = r × dF = r × (K × B)da (22)
Therefore the torque N due to the shell is
N =
r × (K × B)da
= B 0
r × (K × xˆ)da
= B 0
[(r · xˆ)K − xˆ(r · K)] da
= B 0
(x)K − xˆ(r · K 0 sin θ φˆ)
da
= B 0
(x)Kda
= B 0
(a sin θ cos φ)(σωa sin θ φˆ)(a^2 sin θdθdφ) (23)
= B 0 σa^4 ω
∫ (^) π
0
sin^3 θdθ
∫ (^2) π
0
cos φ φdφˆ
= B 0 σa^4 ω
∫ (^) π
0
sin^3 θdθ
∫ (^2) π
0
cos φ(cos φ yˆ − sinφ xˆ)dφ
= B 0 σa^4 ω
∫ (^) π
0
sin^3 θdθ
∫ (^2) π
0
(cos^2 φ yˆ − sin φ cos φ xˆ)dφ
We use
∫ (^2) π 0 sin^ φ^ cos^ φdφ^ = 0,^
∫ (^2) π 0 cos (^2) φdφ = π, and ∫^ π 0 sin (^3) θdθ = 4/3 and get
N =^4 π 3 B 0 σa^4 ω yˆ (24)
(d) Using the formula N = m × B, we get N =^4 π 3 σa^4 ω zˆ × B 0 xˆ =^4 π 3 B 0 σa^4 ω yˆ
