Electrostatics/Electromagnetism

Practice Exam 1 Solutions Notes

Page 1: Electric Forces and Fields

• Problem 1:

  • Electric force vectors: Sum and components
  • Given forces: $Fn$ and $Fl$
  • Volumes for calculation: 15µL and 13µL
  • Force equations:
    • Horizontal component calculation:
    • $Fx = K{9,93} imes ext{cos}(37^{ ext{o}}) = 76.0 ext{N}$
    • Vertical component calculation:
    • $Fy = K{qz qz} imes ext{sin}(37^{ ext{o}}) = 1370 ext{N}$
  • Total Electrostatic Force:
    • Magnitudes: $F_{ ext{total}} = 76.0 ext{N}$ and $1370 ext{N}$

• Problem 2:

  • Electric field definition:
  • The electric field, $E$, at a point in space measures force per unit charge, expressed as:
    • $E = \frac{F}{q}$

• Problem 3:

  • Force on charge in an electric field:
  • Equation:
    • $F_E = qE$
  • Components:
    • $F_x = T ext{sin}( heta)$
    • $F_y = T ext{cos}( heta) = mg$
  • Setting equations equal:
    • $T ext{cos}(0) = mg$
    • Resulting equations:
    • $E = \frac{mg}{T ext{tan}( heta)}$
    • Example calculation yields:
    • $E = 2.40 imes 10^{3} ext{N/C}$

• Problem 4:

  • Electric field dependence:
  • Statement: Magnitude of the electric field does depend on the sign of the charge causing the field.
  • Important equation:
    • $E = k \frac{q}{r^2}$

Page 2: Capacitors and Electric Fields

• Problem 5:

  • Electric fields canceling out:
  • Given charge situation:
    • Total Electric Field, $E_{ ext{net}}$, from charges (considering directions)
    • Calculation yields:
    • $E = 1.80 imes 10^4 ext{N/C}$

• Problem 6:

  • Voltage and capacitance relationship under power:
  • Given the capacitor behavior when connected to power:
    • $Va = K rac{E0 A}{d}$
    • Final charge equation:
    • $Q_{ ext{new}} = 2Q$

• Problem 7:

  • Resistors in parallel:
  • Total resistance decrease leads to increased total current:
  • Ohm's Law:
    • $V = I R$

Page 3: Circuit Analysis

• Problem 8:

  • Equivalent resistance calculation for configurations:
  • Calculate total:
    • $R_{ ext{total}} = R + R = \frac{8}{8} + 16$
    • Yield current:
    • Total current of $1.50 ext{A}$

• Problem 9:

  • Voltage across resistors:
  • For $8 ext{ } ext{Ω}$ resister:
    • $V_g = I R = 1.50 ext{A} imes 8 ext{Ω} = 12.0 ext{V}$
  • Two parallel resistor voltage:
    • $V<em>{ice} = V</em>g = 12.0 ext{V}$

• Problem 10:

  • Current through resistors using Kirchhoff’s laws:
  • $I_{in}$ distributions:
    • Left Loop: Analyze points yielding currents $I_z$, etc.
    • Resulting currents identified through unary equations:
    • $I = 0.67 + 0.15$

Page 4: Magnetic Force and Energy

• Problem 11:

  • Misconception about magnetic forces:
  • False statement: Magnetic force increases speed.
  • Clarification: Magnetic force does NOT increase the speed due to the perpendicular nature to velocity.

• Problem 12:

  • Work-energy theorem application:
  • $W_{ ext{net}} = KE = q imes DV$
  • Formulate final energy values:
    • Combine forces:
    • $F = qvB ext{sin}( heta)$

• Problem 13:

  • Magnetic field relationships:
  • Rule detailing how forces can oppose each other:
    • Resulting calculations on magnetic interaction
    • Derived constants noted.

Page 5: Induced Current and Power Relationships

• Problem 14:

  • Changes in magnetic flux induce currents:
  • Relationship established through defined equations.

• Problem 15:

  • Power input and output relationship:
  • Recognizing values through series definitions:
    • $P<em>{input} = I</em>p V_p = 150A imes 6000V = 9.0 imes 10^{5}W$
    • Establish total power calculations based on resistances:
    • Including efficiency evaluations for outputs at calculated amperage.

• END OF PRACTICE EXAM SOLUTIONS