Electrostatics, Magnetism, and Electromagnetism Problems

Posted on Mar 21, 2025 in Physics

Problem 1: Electric Field, Coulomb Force, and Potential Energy

(a) Diagram of Electric Field Lines

Question:
Two point charges are located in a plane rectangular coordinate system:

• q1 = 10 nC is at (x=0, y=0).
• q2 = 10 nC is at (x=20 cm, y=0).
  Draw a diagram of the electric field lines for these two charges.

Solution:
Since both charges are positive, their electric field lines radiate outward. The field lines repel each other in the space between the charges, bending outward. The key features of the diagram should include:

• Radial outward field lines from each charge.
• Symmetric field distribution along the x-axis.
• A repulsion pattern in the middle region.

(b) Calculate the Coulomb Force on q3

Question:
Calculate the Coulomb force that these two charges exert on a third charge q3 = 1 nC located at (x=5 cm, y=0).

Given Data:

• q1 = 10 nC = 10×10−9 C (at x=0)
• q2 = 10 nC = 10×10−9 C (at x=20 cm)
• q3 = 1 nC = 1×10−9 C (at x=5 cm)
• Coulomb’s constant: k = 9×109 N·m²/C²
• Distances:
  • r13 = 5 cm = 0.05 m (between q1 and q3)
  • r23 = 15 cm = 0.15 m (between q2 and q3)

Step 1: Apply Coulomb’s Law
The force between two charges is given by:

F = k |q1 q2| / r²

Step 2: Compute Individual Forces
The force exerted by q1 on q3:

F13 = (9×109) (10×10−9) (1×10−9) / (0.05)² = 3.6×10−5 N

The force exerted by q2 on q3:

F23 = (9×109) (10×10−9) (1×10−9) / (0.15)² = 4.0×10−6 N

Since both forces act in the positive x-direction, the net force is:

Fnet = F13 – F23 = (3.6×10−5 – 4.0×10−6) = 3.2×10−5 N

Final Answer: Fnet = 3.2×10−5 N in the positive x-direction.

(c) Calculate Potential Energy of q3

Question:
Calculate the potential energy of the point charge q3 using the Coulomb potential.

Formula for Potential Energy:

U = k q3 (q1/r13 + q2/r23)

Substituting Values:

U = (9 × 109) (1 × 10−9) (10 × 10−9 / 0.05 + 10 × 10−9 / 0.15)

U = (9 × 109) (10−9) (10−8 / 5 × 10−2 + 10−8 / 15 × 10−2)

U = (9 × 109) (10−9) (2 × 10−7 + 0.67 × 10−7)

U = (9 × 109) (10−9) × 2.67 × 10−7

U = 2.4 × 10−6 J

Final Answer: The potential energy of q3 is 2.4×10−6 J.

Problem 2: Electron in a Magnetic Field

(a) Determine Required Electric Field

Question:
An electron with speed 106 m/s is moving in the x-direction in a uniform magnetic field B=0.4 T, which is directed along the y-axis. Determine the magnitude and direction of an additional electric field that would prevent deflection.

Step 1: Calculate the Magnetic Force on the Electron
The force due to the magnetic field is:

FB = q v B

where:

• q = -1.6 × 10−19 C (electron charge)
• v = 106 m/s
• B = 0.4 T

FB = (1.6 × 10−19) (106) (0.4) = 6.4 × 10−14 N

Step 2: Set the Electric Force Equal to Magnetic Force
For no deflection, the electric force must cancel the magnetic force:

FE = FB

q E = q v B

E = v B

E = (106) (0.4) = 4 × 105 V/m

Step 3: Determine the Direction

• The magnetic force is in the negative z-direction (by the right-hand rule).
• The electric field must be in the positive z-direction to cancel it.

Final Answer: The required electric field is 4.0×105 V/m in the positive z-direction.

Problem 3: Self-Inductance and Induced Voltage in a Solenoid

(a) Determine Self-Inductance of a Solenoid

Question:
A circular cylindrical solenoid has:

• Number of turns: N = 1000
• Length: l = 25 cm = 0.25 m
• Diameter: d = 2 cm = 0.02 m

Determine its self-inductance.

Step 1: Use the Formula for Self-Inductance

L = μ0 μr N² A / l

where:

• μ0 = 4π × 10−7 H/m (permeability of free space)
• μr = 1 (assuming air core)
• A = π r² = π (0.01)² = 3.14 × 10−4 m²

Step 2: Compute the Inductance

L = (4π × 10−7) × (1000)² × (3.14 × 10−4) / 0.25

L = (4π × 10−7) × (4 × 106) × (3.14 × 10−4) ÷ 0.25

L = 1.57 × 10−3 H = 1.57 mH

Final Answer: The self-inductance is 1.57 mH.

(b) Calculate Induced Voltage in a Wire Loop

Question:
A wire loop with 1 turn and a diameter of 5 cm is placed concentrically around the solenoid.
The solenoid current varies as:

I = 1.25 A sin(ωt)

where ω = 2πf = 2π × 50 = 314.16 rad/s.

Find the induced voltage V(t) in the loop.

Step 1: Find Magnetic Flux

Magnetic field inside solenoid:

B = μ0 (N/l) I

B = (4π × 10−7) (1000/0.25) (1.25 sin(314.16 t))

B = (4π × 10−7 × 4000) (1.25 sin(314.16 t))

B = (0.005) (1.25 sin(314.16 t))

B = 0.00625 sin(314.16 t) T

Flux through the wire loop:

Φ = B A

Φ = (0.00625 sin(314.16 t)) (π (0.025)²)

Φ = (0.00625 sin(314.16 t)) (1.96 × 10−3)

Φ = 1.23 × 10−5 sin(314.16 t) Wb

Step 2: Find Induced EMF (Faraday’s Law)

V = – dΦ/dt

V = – (1.23 × 10−5) (314.16 cos(314.16 t))

V = -3.87 × 10−3 cos(314.16 t) V

Final Answer: The induced voltage is −3.87 mV cos(314.16 t).

Problem 4: Electromagnetic Spectrum

Question:
Name four other types of waves or radiation in the electromagnetic spectrum and their approximate wavelength or frequency ranges.

Problem 5: Capacitor with Dielectric

(a) Calculate Charge Stored in Capacitor

Question:
A capacitor has:

• Plate spacing: d = 0.1 mm = 1×10−4 m
• Plate area: A = 1.5 cm² = 1.5×10−4 m²
• Voltage: U = 10 V
• Dielectric constant: εr = 50

Find the charge on each plate.

Step 1: Calculate Capacitance

C = ε0 εr A / d

C = (8.85 × 10−12 × 50) × 1.5 × 10−4 / 10−4

C = 6.64 × 10−11 F

Step 2: Calculate Charge

Q = C U = (6.64 × 10−11) (10)

Q = 6.64 × 10−10 C

Final Answer: The charge is 664 pC.

(b) Find Stored Energy

U = 1/2 C V²

U = 1/2 (6.64 × 10−11) (10)²

U = 3.32 × 10−9 J

(c) How Does Energy Change When Dielectric is Removed?

Since C reduces to:

C′ = C / εr = 6.64 × 10−11 / 50

C′ = 1.33 × 10−12 F

Stored energy:

U′ = Q² / (2 C′)

U′ = (6.64 × 10−10)² / (2 (1.33 × 10−12))

U′ = 1.66 × 10−7 J

Energy increases by a factor of 50.

(d) How Does Capacitor Voltage Change?

V′ = Q / C′ = 6.64 × 10−10 / 1.33 × 10−12

V′ = 500 V

Voltage increases from 10V to 500V.

Problem 6: Reduction in Electric and Magnetic Field Strengths

Question:
Sunglasses have a tint of 75%, meaning they transmit only 25% of the incident light intensity.
By what percentage are the electric field (E) and magnetic field (B) strengths reduced?

Step 1: Relationship Between Intensity and Field Strengths

The intensity of an electromagnetic wave is proportional to the square of the electric field strength:

I ∝ E²

Since the magnetic field B is related to the electric field in an electromagnetic wave, the same relationship holds:

I ∝ B²

Step 2: Calculate the Reduction Factor

Given that the transmitted intensity is 25% of the incident intensity:

I′ = 0.25 I

Taking the square root on both sides:

E′ = E √0.25 = E × 0.5

B′ = B √0.25 = B × 0.5

Thus, both the electric field and magnetic field strengths are reduced to 50% of their original values.

Step 3: Percentage Reduction

The reduction in field strength is:

100% – 50% = 50%

Final Answer:
The electric and magnetic field strengths are reduced by 50% as the light passes through the sunglasses.

Here are the questions in English, exactly as given, along with their answers and steps.

Problem 7: True or False Statements

(a) Electric Field of a Point Charge

Question: The electric field of a point charge always points away from the charge.
Answer: False.

• The electric field direction depends on the sign of the charge.
• For a positive charge, the field points away.
• For a negative charge, the field points toward the charge.

(b) Macroscopic Charges

Question: All macroscopic charges q can be written as q = ± ne, where n is an integer and e is the elementary charge.
Answer: True.

• Charge is quantized, meaning any charge is an integer multiple of the elementary charge e = 1.602 × 10−19 C.
• This is a fundamental principle of quantum mechanics.

(c) Electric Field Lines Divergence

Question: Electric field lines never diverge from a point in space.
Answer: False.

• Electric field lines originate from positive charges and terminate at negative charges.
• If there is a positive charge at a point, field lines diverge from it.
• If there is a negative charge at a point, field lines converge to it.

(d) Electric Field Lines Crossing

Question: Electric field lines never cross at a point in space.
Answer: True.

• If two electric field lines were to cross, it would mean that at that point, there are two different directions for the electric field, which is not possible.
• The electric field at any point in space has a unique direction.

(e) Electric Dipole Moments

Question: All molecules have electric dipole moments in the presence of an external electric field.
Answer: False.

• Only polar molecules (e.g., water) have a permanent dipole moment.
• Non-polar molecules (e.g., oxygen, nitrogen) do not have a dipole moment unless an electric field induces one.

Problem 8: KCl Ion Separation Energy

Question: The K+ and the Cl− ion have a distance of 2.08 × 10−10 m in the KCl. Calculate the energy that must be expended to separate the two ions to an infinite distance. Assume that they are point particles that are initially at rest. Give the result in eV.

Answer:
The electrostatic potential energy between two point charges is given by:

U = k q1 q2 / r

where:

• k = 8.99 × 109 Nm²/C² (Coulomb’s constant),
• q1 = e = 1.602 × 10−19 C (charge of K+),
• q2 = -e = -1.602 × 10−19 C (charge of Cl−),
• r = 2.08 × 10−10 m.

Substituting the values:

U = (8.99 × 109) (1.602 × 10−19)² / (2.08 × 10−10) = 6.92 × 10−19 J

To convert to electron volts (eV):

1 eV = 1.602 × 10−19 J

U = (6.92 × 10−19) / (1.602 × 10−19) = 4.32 eV

Thus, 4.32 eV of energy is required to separate the ions to infinity.

Problem 9: Capacitor Energy Changes

(a) Constant Voltage

Question: An air-filled plate capacitor is connected to a battery with constant voltage. How does the energy stored in the capacitor change if the distance between the capacitor plates is doubled while the battery is still connected?

Answer:
The energy stored in a capacitor is given by:

U = 1/2 C V²

where:

• C = ε0 A / d (capacitance formula),
• d is the plate separation.

If d is doubled (d′ = 2d), the new capacitance becomes:

C′ = ε0 A / (2d) = C / 2

Since the battery remains connected, voltage V is constant.

U′ = 1/2 C′ V² = 1/2 × (C/2) × V² = U/2

Thus, the energy is halved.

(b) Disconnected Capacitor

Question: How does the energy stored in the capacitor of task (a) change if it is disconnected from the battery before the distance between the plates is doubled?

Answer:
If the capacitor is disconnected, the charge Q is constant instead of voltage.

Since:

U = Q² / (2C)

and capacitance is halved when d is doubled:

C′ = C/2

U′ = Q² / (2(C/2)) = 2U

Thus, the stored energy doubles.

Problem 10: Mean Current and Magnetic Dipole Moment

(a) Mean Current

Question: Show that the mean value of the current is given by I = qω / (2π).

Solution:
The charge q moves in a circular path with angular velocity ω.
Current is defined as charge per unit time:

I = charge per cycle / time per cycle

• The charge completes one full revolution in a period:

T = 2π / ω

• The charge passing per revolution is q.

I = q/T = q / (2π/ω)

I = qω / (2π)

Thus, the mean current is I = qω / (2π).

(b) Magnetic Dipole Moment

Question: Show that the magnitude of the magnetic dipole moment is mmag = 1/2 qωr².

Using mmag = IA, where A = πr²,

mmag = (qω / (2π)) × (πr²)

mmag = 1/2 qωr²

Thus, proven.

Problem 11: Magnetic Field of Infinite Wires

Question: An infinitely long, insulated wire lies on the x-axis of a coordinate system and a current I flows through it in the positive x-direction. A second, similar wire lies on the y-axis and the current I flows through it in the positive y-direction. At which point or points in the x-y plane is the resulting magnetic field zero?

Solution:

• The magnetic field due to an infinite wire carrying current I is given by Ampère’s Law: B = μ0 I / (2πr) where r is the perpendicular distance from the wire.
• The right-hand rule gives the direction of the field:
  • The x-axis wire creates a field that circulates around it.
  • The y-axis wire does the same.
• We need to find points where these fields cancel.

Step-by-step Analysis:

1. Magnetic field due to wire along x-axis:

• The field at a point (x,y) is given by: Bx = μ0 I / (2πy)
• Direction: Counterclockwise when viewed from positive x-axis.

Magnetic field due to wire along y-axis:

• The field at a point (x,y) is given by: By = μ0 I / (2πx)
• Direction: Counterclockwise when viewed from positive y-axis.

Setting the two fields equal for cancellation:

• The only possible points where the fields might cancel are where their magnitudes are equal but in opposite directions.
• This happens when: μ0 I / (2πy) = μ0 I / (2πx)
• Simplifying: x = y

Conclusion:

• The points where the magnetic field is zero lie along the line y = x, but not at the origin (0,0), where the fields are undefined.

Problem 12: Laser Pulse Properties

(a) Spatial Length of Pulse

Question: A laser pulse with an energy of 20 J and a beam radius of 2 mm lasts for 10 ns. The energy density during the pulse is constant.

Solution:

• The speed of light in vacuum is c = 3.00 × 108 m/s.
• The spatial length L of the pulse is: L = c × t where t = 10 ns = 10 × 10−9 s.

L = (3.00 × 108 m/s) × (10 × 10−9 s) = 3.00 m

Thus, the spatial length of the pulse is 3.00 m.

(b) Energy Density

Solution:

• Energy density u is the energy per unit volume: u = E/V where:
  • E = 20 J (total energy),
  • Volume V = A × L,
  • A (cross-sectional area of beam) = πr², where r = 2 mm = 2 × 10−3 m.

A = π (2 × 10−3)² = π × 4 × 10−6 = 1.26 × 10−5 m²

V = (1.26 × 10−5) × (3.00) = 3.77 × 10−5 m³

u = 20 / (3.77 × 10−5) = 5.31 × 105 J/m³

Thus, the energy density is 5.31 × 105 J/m³.

(c) Electric and Magnetic Field Strengths

Solution:

1. Electric field strength E is related to energy density by:

u = 1/2 ε0 E²

Solving for E:

E = √(2u / ε0)

Using ε0 = 8.85 × 10−12 F/m:

E = √(2 (5.31 × 105) / (8.85 × 10−12))

E = √(1.2 × 1017)

E = 1.1 × 109 V/m

Thus, E = 1.1 × 109 V/m.

1. Magnetic field strength B is related to E by:

B = E/c

B = (1.1 × 109) / (3.00 × 108) = 3.67 T

Thus, B = 3.67 T.