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Derive the mathematical expression for electric field intensity and solve problems related to parallel capacitors.

Physics

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Here are the solutions to questions Q.14, Q.15, and Q.16.

Q.14:

a)

  • Coulomb's Law: Coulomb's Law states that the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The force acts along the line joining the two charges.
  • Mathematical Expression for Electric Field Intensity:
  1. State Coulomb's Law for the force FF between a point charge qq and a test charge q0q_0 at a distance rr.

    $$
    F = \frac{1}{4\pi\epsilon_0} \frac{q q_0}{r^2}
    $$
    
  2. Define electric field intensity EE as the force per unit test charge (E=F/q0E = F/q_0).

    $$
    E = \frac{F}{q_0} = \frac{1}{q_0} \left( \frac{1}{4\pi\epsilon_0} \frac{q q_0}{r^2} \right)
    $$
    
  3. Simplify the expression.

    $$
    E = \boxed{\frac{1}{4\pi\epsilon_0} \frac{q}{r^2}}
    $$
    

    b)
    Given: C1=10μFC_1 = 10 \mu\text{F}, C2=20μFC_2 = 20 \mu\text{F}, V=90 VV = 90 \text{ V}.

    a) Calculate the equivalent capacitance of the combination.

  4. For capacitors connected in parallel, the equivalent capacitance is the sum of individual capacitances.

    Ceq=C1+C2C_{eq} = C_1 + C_2
  5. Substitute the given values.

    Ceq=10μF+20μFC_{eq} = 10 \mu\text{F} + 20 \mu\text{F} Ceq=30μFC_{eq} = \boxed{30 \mu\text{F}}

    b) Calculate the charge stored on each capacitor.

  6. In a parallel connection, the voltage across each capacitor is the same as the source voltage (V=90 VV = 90 \text{ V}).

    Use the formula Q=CVQ = CV.
    For C1C_1:

    Q1=C1VQ_1 = C_1 V Q1=(10×10−6 F)×(90 V)Q_1 = (10 \times 10^{-6} \text{ F}) \times (90 \text{ V}) Q1=900×10−6 C or 900μCQ_1 = \boxed{900 \times 10^{-6} \text{ C} \text{ or } 900 \mu\text{C}}

    For C2C_2:

    Q2=C2VQ_2 = C_2 V Q2=(20×10−6 F)×(90 V)Q_2 = (20 \times 10^{-6} \text{ F}) \times (90 \text{ V}) Q2=1800×10−6 C or 1800μCQ_2 = \boxed{1800 \times 10^{-6} \text{ C} \text{ or } 1800 \mu\text{C}}

    Q.15:

    a)

    • Ampere's Law: Ampere's Law states that the line integral of the magnetic field B⃗\vec{B} around any closed loop is equal to μ0\mu_0 times the total current IencI_{enc} passing through the loop. ∮B⃗⋅dl⃗=μ0Ienc\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{enc}
    • Magnetic Field Inside an Ideal Solenoid:
  7. Consider an ideal solenoid with nn turns per unit length carrying current II.

    The magnetic field inside is uniform and parallel to the axis, and zero outside.

  8. Choose a rectangular Amperian loop ABCD of length LL inside the solenoid, with side AB parallel to the axis and side CD outside.

  9. Apply Ampere's Law.

    The line integral ∮B⃗⋅dl⃗\oint \vec{B} \cdot d\vec{l} simplifies to BLBL because the field is uniform and parallel to AB, and zero or perpendicular along other segments. ∮B⃗⋅dl⃗=BL \oint \vec{B} \cdot d\vec{l} = BL

  10. Calculate the total current enclosed by the loop.

    If there are nn turns per unit length, then in length LL, there are nLnL turns. Each turn carries current II. Ienc=nLI I_{enc} = nLI

  11. Equate the line integral and μ0Ienc\mu_0 I_{enc}.

    $$
    BL = \mu_0 nLI
    $$
    
  12. Solve for BB.

    $$
    B = \boxed{\mu_0 n I}
    $$
    

    b)
    The four factors that affect the magnitude of the force acting on a charged particle moving in a magnetic field are:

    • The magnitude of the charge (qq) of the particle.
    • The magnitude of the velocity (vv) of the particle.
    • The magnitude of the magnetic field strength (BB).
    • The angle (θ\theta) between the velocity vector of the particle and the magnetic field vector.

    Q.16:

    a)

    • Condition for Resonance: In an RLC series AC circuit, resonance occurs when the inductive reactance (XLX_L) is equal to the capacitive reactance (XCX_C). XL=XCX_L = X_C
    • Formula for Resonant Frequency:
  13. Express reactances in terms of frequency.

    $$
    \omega_0 L = \frac{1}{\omega_0 C}
    $$
    
  14. Solve for the angular resonant frequency ω0\omega_0.

    $$
    \omega_0^2 = \frac{1}{LC} \implies \omega_0 = \frac{1}{\sqrt{LC}}
    $$
    
  15. Convert angular frequency to linear frequency (f0=ω0/2πf_0 = \omega_0 / 2\pi).

    $$
    f_0 = \boxed{\frac{1}{2\pi\sqrt{LC}}}
    $$
    
    • Impedance at Resonance:
  16. The impedance ZZ of an RLC series circuit is given by Z=R2+(XL−XC)2Z = \sqrt{R^2 + (X_L - X_C)^2}.

  17. At resonance, XL=XCX_L = X_C, so (XL−XC)=0(X_L - X_C) = 0.

    $$
    Z = \sqrt{R^2 + 0^2}
    $$
    $$
    Z = \boxed{R}
    $$
    

    b)
    Given: Vrms=200 VV_{rms} = 200 \text{ V}, Irms=0.5 AI_{rms} = 0.5 \text{ A}, power factor cos⁡ϕ=0.6\cos\phi = 0.6.

  18. Use the formula for average power consumed in an AC circuit.

    P=VrmsIrmscos⁡ϕP = V_{rms} I_{rms} \cos\phi
  19. Substitute the given values.

    P=(200 V)×(0.5 A)×(0.6)P = (200 \text{ V}) \times (0.5 \text{ A}) \times (0.6)
  20. Calculate the power.

    P=100 W×0.6P = 100 \text{ W} \times 0.6 P=60 WP = \boxed{60 \text{ W}}
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