More Problems

Problem 1:

Three charges, -Q, -Q and +2Q, where Q = 1 microCoulomb, are arranged on an equilateral triangle with sides of length L = 1 m as shown.  Point M is on the midpoint of the line joining the two negative charges.  Potentials are referenced to zero at infinity.

(a)  Compute the force (magnitude with correct units and direction) on the charge +2Q due to the other two charges.
(b)  Compute the electric field (magnitude with correct units and direction) at point M produced by the three charges.
(c)  Calculate the energy necessary for bringing a +Q charge from infinity to point M.
(d)  Calculate the total energy necessary to assemble the whole system of four charges.

Solution:

• Concepts:
  Coulomb's law, the principle of superposition, electrostatic energy
• Reasoning:
  We are asked to answer questions concerning a system of point charges.
• Details of the calculation:
  Let the x-axis point towards the right and the y-axis upward.
  (a)  F(2Q) = (-j) (4kQ²/L²) cos(30^o) = -3.464 kQ²/L² j = -3.12*10^-2 N j.
  (b)  E(M) = (-j) 2kQ/(L cos(30^o))² (-j) = -2.667 kQ/L² j = -2.4*10⁴ N/C j.
  (c)  U = 2kQ²/(L cos(30^o)) - 2kQ²/(L sin(30^o)) = -1.69 kQ²/L = -1.52*10^-2 J.
  (d)  The energy necessary to assemble the 3 original charges is
  U = -4kQ²/L + kQ²/L = -3kQ²/L = -2.7*10^-2 J.
  Therefore the energy necessary to assemble the 4 charges is
  U = -2.7*10^-2 J - 1.52*10^-2 J = -4.22*10^-2 J.

Problem 2:

This question probes your understanding of dielectrics and associated fields and sources.  For this question, dielectric means linear isotropic homogeneous (lih) dielectric.
(a)  If one presumes that there exists a true charge density ρ_true, a polarization or bound charge density ρ_bound, and a total charge density ρ_total, such that ρ_true + ρ_bound = ρ_total, write the source equations for D, E, and P.  Explain the meaning of these equations.  Briefly address the question:  Which of the fields D or E might be considered the more fundamental field?  Why?  Write the equation(s) describing the relationships between the three field quantities.

(b)  Draw a diagram of a lih dielectric of thickness c between the plates of a parallel plate capacitor of separation d with a gap of thickness e between each plate and the dielectric.  Presume that there is a constant voltage of V volts applied to the capacitor at time t = 0 (by, for example, connecting a V volt battery at time t = 0.)  Discuss and draw diagrams of the fields at t = dt for very small dt and the evolution of the fields to t = ∞.

(c)  Repeat part (b) with the lih dielectric replaced by a conductor.  Make sure that you consider all of the fields, creatively defining P in the conductor, masquerading (for purposes of this question) as a dielectric.  Comment on the similarities and differences between the lih dielectric and the conductor as a dielectric.  (It is strongly suggested that you do not consider part (c) as meaningless.)

Solution:

• Concepts:
  Maxwell's equations for electrostatics
• Reasoning:
  Maxwell's equations connect the source equations for E and P and yield the source equations for D.
• Details of the calculation:
  (a)  Maxwell's equations for electrostatics are
  ∇∙E = ρ_total/ε₀,  ∇×E = 0, (SI units).
  ∇∙E = ρ_total/ε₀ is the source equation for E.
  Let ρ_true + ρ_bound = ρ_total, then ∇∙E = (ρ_true + ρ_bound)/ε₀ = ρ_true/ε₀ - ∇∙P/ε₀.
  ∇∙P = -ρ_bound is the source equation for P.
  ∇∙(ε₀E + P) = ρ_true.  D = ε₀E + P, ∇∙D = ρ_true is the source equation for D.
  Space derivatives of E, P, and D must exist, otherwise use the integral form.
  For example:   ∫_V ∇∙D dV = ∮_A D∙n dA = ∫_V ρ_true dV = Q_{free inside}.
  For lih materials:
  P = ε₀χ_eE, D = ε₀(1 + χ_e)E =  ε₀κ_eE = εE.
  Equations of macroscopic electrostatics:  ∇∙D = ρ_true,  ∇×E = 0.
  D should not be regarded as a fundamental field of the same status as E.  It is rather a mathematical construct related to the way in which we seek a macroscopic solution for E from the basic equations.  D is a useful construct if the material has special properties, for example if it is a lih dielectric.  Then D = εE.  Unfavorable external conditions, however, can destroy those special properties.
  
  (b)  This is a quasi-static problem.  The voltage across the plates is constant.  For a very short time a polarization current is flowing, until the free and bound surface charge densities have reached their equilibrium values.
  The magnitude of the electric field produced by two large plates with surface charge densities ±½σ, separated by a small distance d, is E = V/d = σ/ε₀ in the region between the plates.  For the plates in this problem, at any time t, D is uniform throughout the capacitor.  D, E and P point in the x-direction.  E abruptly changes at the vacuum-dielectric boundary.  P is only non-zero in the dielectric.
  Let j_p denote the polarization current, σ_f the magnitude free surface charge density on the capacitor plates, and σ_b the magnitude of the bound surface charge density on the dielectric.
  At any time we have V = (σ_f/ε₀)d - (σ_b/ε₀)c,
  σ_b = |P∙n|,  dσ_b/dt = dP/dt = j_p,  σ_f = D.
  At t = 0, σ_b = 0 and P = 0.  E₀ = V/d inside the dielectric, and σ_f = ε₀V/d.
  As t --> ∞, σ_b = P = D - ε₀E = D(1 - ε₀/ε) = σ_f(1 - ε₀/ε).
  Therefore σ_f = (ε₀V + σ_bc)/d = ε₀V/(d + c(ε₀/ε - 1)).
  Inside the dielectric E = D/ε = σ_f/ε = ε₀V/(εd + c(ε₀ - ε )).
  As we connect the battery, we can assume that the circuit behaves like a RC circuit, and that σ_b increases from 0 to σ_b(∞).
  σ_b(t) = σ_b(∞)(1 - e^-t/τ), where τ is the time constant of the circuit.  For very small times dt, σ_b(t) = σ_b(∞)dt/τ increases linearly.
  σ_f(t) = (ε₀V + σ_b(t)c)/d also increases with time.
  The electric field in the vacuum region E_out(t) = V/d + σ_b(t)c/(dε₀) increases with time, and the electric field in the dielectric E_in(t) = (V - E_out(t)(d - c))/c decreases with time.
  
  (c)  In the conductor, as t --> ∞, E = 0.  But for very small times E = j/σ_c, where σ_c is the conductivity.
  Under certain circumstances, when no steady currents can flow and the free charges cannot travel large distances, we can formally treat the conductor like a dielectric with a very large ε and only bound charges.
  For inside the conductor we the write dP/dt = dσ_b/dt = j_p = σ_cE_in.
  For a conductor with a given conductivity we can find the time constant τ.
  We have P = D - ε₀E_in = σ_b = σ_f -  ε₀j/σ_c.
  σ_b = σ_f - (ε₀/σ_c)dσ_b/dt.
  dσ_b/dt = (σ_c/ε₀)(σ_f - σ_b) =  (σ_c/ε₀)((ε₀V + σ_b(t)c)/d - σ_b(t))
  = σ_cV/d - (1 - c/d)(σ_c/ε₀)σ_b.
  σ_b(t) = (ε₀V/(d - c))(1 - e^-t/τ) = σ_f(∞)(1 - e^-t/τ), with 1/τ = (1 - c/d)(σ_c/ε₀).
  For very small times dt, σ_b(t) increases linearly.  σ_f(t) also increases with time.
  The electric field in the vacuum region increases from V/d to V/(d - c), and the electric field in the conductor decreases from V/d to zero.

Problem 3:

An ideal battery is connected to a 200 kΩ and a 300 kΩ resistor in series.
A voltmeter is used to measure the voltage across the battery and the 200 kΩ resistor.
Its readings are V_battery = 6.0 V,  V_200kΩ = 2 V.
What is the value is the internal resistance of the voltmeter and what will the voltmeter read if it is used to measure the voltage across the 300 kΩ resistor?

Solution:

• Concepts:
  Resistors in series and parallel
• Reasoning:
  The voltmeter as has a shunt resistance R.  It reads the voltage across this resistance.  When placed in a circuit, it correctly reads the battery voltage but not the voltages across the resistors of the undisturbed circuit.  The battery voltage is V = 6 V.
  The voltages across R₁ (200 kΩ) and R₂ (300 kΩ) (without the voltmeter in the circuit) are
  V₁ = 6V R₁/(R₁ + R₂) = 2.4 V and V₂ = 6V R₂/(R₁ + R₂) = 3.6 V.
• Details of the calculation:
  With the voltmeter in the circuit we have:
  2V = 6V [R₁R/(R₁ + R)]/[(R₁R/(R₁ + R)) + R₂] or 2V = 6V/[1 + R₂(R₁ + R)/(R₁R)].
  1 + R₂(R₁ + R)/(R₁R)] = 3,  R₂(R₁ + R)/(R₁R) = 2,  R₂/R + R₂/R₁ = 2,  1/R = 2/R₂ - 1/R₁
  R = 600 kΩ is the internal resistance of the voltmeter.
  When placed across the 300 kΩ resistor the voltmeter reading is
  V_300kΩ = 6V [R₂R/(R₂ + R)]/[(R₂R/(R₂ + R)) + R₁] = 3 V.

Problem 4:

A cylindrical capacitor of length L has an inner cylinder of radius a and an outer cylinder of radius b.  A potential difference V₀ is maintained between the cylinders.
(a)  Find the magnitude of the electric field E(r) in the region a < r < b in terms of V₀, a, and b, assuming L >> a, b.
(b)  Find the capacitance in terms of L, a, and b.
(c)  A dielectric with permittivity ε is now inserted.  It fills the space between the cylinders.  Find the change in the capacitance.
(d)  How much work must an external force do to pull out (remove) the dielectric?

Solution:

• Concepts:
  Gauss' law
• Reasoning:
  The problem has enough symmetry to find E(r) from Gauss' law alone.
• Details of the calculation:
  (a)  From Gauss' law: E(r ) ∝ 1/r between the cylinders.
  Let the outer cylinder be grounded and V₀ be positive.  Then the direction of E is radially outward. 
  E(r) = A/r, with A positive. 
  V₀ = -∫_b^aE(r)dr = A∫_a^b(1/r)dr = A(lnb - lna) = A ln(b/a).  A = V₀/ln(b/a).
  E(r) = V₀/(r ln(b/a)).  E(a) = V₀/(a ln(b/a)).
  (b)  C = Q/V₀ = ε₀E(a)2πaL/V₀ = 2πε₀L/(ln(b/a)) = -2πε₀L/ln(a/b).
  (c)  C_new = -2πεL/ln(a/b).  ΔC = C_new - C_initial = -2π(ε - ε₀)L/ln(a/b).
  (d)  The work done by an external force as a dielectric is pulled out is W = -∫F_xdx, assuming the axis of the cylinders is the x-axis.
  F_x = -dW_mech/dx.  F_x is the force with which the electric force pulls on the dielectic.
  The capacitance as a function of x, the distance the dielectric has moved as it is being pulled out.
  C = 2πε₀x/(ln(b/a)) + 2πε(L - x)/(ln(b/a)).
  Total energy store in the capacitor:  U = ½CV₀².
  dW_mech + dW_bat = dU.  dW_bat = dQV₀ = dCV₀².  dU = ½dCV₀². 
  dW_mech = -½dCV₀². 
  F_x = -dW_mech/dx = ½V₀²dC/dx  = -½V₀²2π(ε - ε₀)/(ln(b/a).
  The electric force is independent of x.  It is pulling the dielectric into the capacitor.
  W_mech = V₀²π(ε - ε₀)L/(ln(b/a) is the work required to remove the dielectric.

Problem 5:

A dipole p = pk = 4*10^-12 Cm is located at the origin. 
Find the force it exerts on a point charge q = 5*10^-6 C located on the z-axis at z = 10^-4 m.

Solution:

• Concepts:
  The dipole field,  E(r) = [1/(4πε₀)](1/r³)[3(p∙r)r/r² - p], F = qE.
• Reasoning:
  E = (p/(4πε₀r³))[2cosθ e_r + sinθ e_θ'].
  On the z-axis E = (p/(2πε₀r³)) k.
• Details of the calculation:
  F = 5*10^-6*4*10^-12*4.5*10⁹/10^-12 N.
  F = 9*10⁴ N k.