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Identify given values and the principle for equilibrium.

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5 parts
a

60μF\text{}60 \mu\text{F}

b

60.03μF\text{}60.03 \mu\text{F}

b(i)

0.4 A\text{}0.4 \text{ A}

b(ii)

10Ω\text{}10 \Omega

b(iii)

12 V\text{}12 \text{ V}

Step-by-step solution

13 steps

Step 1: Identify given values and the principle for equilibrium.
The oil drop is stationary, meaning the upward electric force (FeF_e) balances the downward gravitational force (FgF_g).
Given:
Mass, m=5×10−15 kgm = 5 \times 10^{-15} \text{ kg}
Acceleration due to gravity, g=10 N/kgg = 10 \text{ N/kg}
Distance between plates, d=25 mm=0.025 md = 25 \text{ mm} = 0.025 \text{ m}
Potential difference, V=1000 VV = 1000 \text{ V}

Step 2: Calculate the gravitational force.

Fg=mgF_g = mg Fg=(5×10−15 kg)(10 N/kg)F_g = (5 \times 10^{-15} \text{ kg})(10 \text{ N/kg}) Fg=5×10−14 NF_g = 5 \times 10^{-14} \text{ N}

Step 3: Calculate the electric field between the plates.

E=VdE = \frac{V}{d} E=1000 V0.025 mE = \frac{1000 \text{ V}}{0.025 \text{ m}} E=40000 V/mE = 40000 \text{ V/m}

Step 4: Equate electric force and gravitational force to find the charge.

Fe=FgF_e = F_g qE=mgqE = mg q=mgEq = \frac{mg}{E} q=5×10−14 N40000 V/mq = \frac{5 \times 10^{-14} \text{ N}}{40000 \text{ V/m}} q=5×10−144×104 Cq = \frac{5 \times 10^{-14}}{4 \times 10^4} \text{ C} q=1.25×10−18 Cq = 1.25 \times 10^{-18} \text{ C}

The charge on the drop is 1.25×10−18 C\boxed{\text{}1.25 \times 10^{-18} \text{ C}}.

Part a· 2 steps

The capacitance of C1C_1 and C2C_2 in series.

  1. Identify given values and the formula for the time constant.

    Given:
    Resistance, R=80 kΩ=80×103ΩR = 80 \text{ k}\Omega = 80 \times 10^3 \Omega
    Time constant, τ=4.8 s\tau = 4.8 \text{ s}
    The time constant for an RC circuit is given by τ=RCeq\tau = RC_{eq}, where CeqC_{eq} is the equivalent capacitance.

  2. Calculate the equivalent capacitance CeqC_{eq}.

    τ=RCeq\tau = RC_{eq} Ceq=τRC_{eq} = \frac{\tau}{R} Ceq=4.8 s80×103ΩC_{eq} = \frac{4.8 \text{ s}}{80 \times 10^3 \Omega} Ceq=6×10−5 FC_{eq} = 6 \times 10^{-5} \text{ F} Ceq=60×10−6 F=60μFC_{eq} = 60 \times 10^{-6} \text{ F} = 60 \mu\text{F}

    The capacitance of C1C_1 and C2C_2 in series is 60μF\boxed{\text{}60 \mu\text{F}}.

Part b

60.03μF\text{}60.03 \mu\text{F}

The capacitance of C1C_1 if C2C_2 has a capacitance of 100 mF100 \text{ mF}.

  1. Identify given values and the formula for capacitors in series.

    Given:
    Equivalent capacitance, Ceq=6×10−5 FC_{eq} = 6 \times 10^{-5} \text{ F} (from part a))
    Capacitance of C2=100 mF=100×10−3 F=0.1 FC_2 = 100 \text{ mF} = 100 \times 10^{-3} \text{ F} = 0.1 \text{ F}
    For capacitors in series, the equivalent capacitance is given by:

    1Ceq=1C1+1C2\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}

  2. Solve for C1C_1.

    1C1=1Ceq−1C2\frac{1}{C_1} = \frac{1}{C_{eq}} - \frac{1}{C_2} 1C1=16×10−5 F−10.1 F\frac{1}{C_1} = \frac{1}{6 \times 10^{-5} \text{ F}} - \frac{1}{0.1 \text{ F}} 1C1=16666.666... F−1−10 F−1\frac{1}{C_1} = 16666.666... \text{ F}^{-1} - 10 \text{ F}^{-1} 1C1=16656.666... F−1\frac{1}{C_1} = 16656.666... \text{ F}^{-1} C1=116656.666... FC_1 = \frac{1}{16656.666...} \text{ F} C1≈6.003×10−5 FC_1 \approx 6.003 \times 10^{-5} \text{ F} C1≈60.03μFC_1 \approx 60.03 \mu\text{F}

    The capacitance of C1C_1 is 60.03μF\boxed{\text{}60.03 \mu\text{F}}.

Part b(i)

0.4 A\text{}0.4 \text{ A}

The current in the 6Ω6 \Omega resistor.

  1. Identify given values and the circuit configuration.

    Given:
    Current in 3Ω3 \Omega resistor, I3Ω=0.8 AI_{3\Omega} = 0.8 \text{ A}
    The 3Ω3 \Omega and 6Ω6 \Omega resistors are connected in parallel. In a parallel circuit, the voltage across each branch is the same.

  2. Calculate the voltage across the 3Ω3 \Omega resistor.

    Using Ohm's Law, V=IRV = IR:

    V3Ω=I3Ω×R3ΩV_{3\Omega} = I_{3\Omega} \times R_{3\Omega} V3Ω=0.8 A×3ΩV_{3\Omega} = 0.8 \text{ A} \times 3 \Omega V3Ω=2.4 VV_{3\Omega} = 2.4 \text{ V}

  3. Calculate the current in the 6Ω6 \Omega resistor.

    Since the resistors are in parallel, V6Ω=V3Ω=2.4 VV_{6\Omega} = V_{3\Omega} = 2.4 \text{ V}.

    I6Ω=V6ΩR6ΩI_{6\Omega} = \frac{V_{6\Omega}}{R_{6\Omega}} I6Ω=2.4 V6ΩI_{6\Omega} = \frac{2.4 \text{ V}}{6 \Omega} I6Ω=0.4 AI_{6\Omega} = 0.4 \text{ A}

    The current in the 6Ω6 \Omega resistor is 0.4 A\boxed{\text{}0.4 \text{ A}}.

Part b(ii)

10Ω\text{}10 \Omega

The internal resistance, rr.

  1. Calculate the total current flowing from the battery.

    The total current (ItotalI_{total}) is the sum of the currents in the parallel branches.

    Itotal=I3Ω+I6ΩI_{total} = I_{3\Omega} + I_{6\Omega} Itotal=0.8 A+0.4 AI_{total} = 0.8 \text{ A} + 0.4 \text{ A} Itotal=1.2 AI_{total} = 1.2 \text{ A}

  2. Calculate the equivalent resistance of the parallel combination.

    For parallel resistors:

    1Rp=1R3Ω+1R6Ω\frac{1}{R_p} = \frac{1}{R_{3\Omega}} + \frac{1}{R_{6\Omega}} 1Rp=13Ω+16Ω\frac{1}{R_p} = \frac{1}{3 \Omega} + \frac{1}{6 \Omega} 1Rp=26Ω+16Ω\frac{1}{R_p} = \frac{2}{6 \Omega} + \frac{1}{6 \Omega} 1Rp=36Ω\frac{1}{R_p} = \frac{3}{6 \Omega} Rp=63Ω=2ΩR_p = \frac{6}{3} \Omega = 2 \Omega

  3. Calculate the total external resistance.

    The parallel combination (RpR_p) is in series with the 8Ω8 \Omega resistor.

    Rext=Rp+8ΩR_{ext} = R_p + 8 \Omega Rext=2Ω+8ΩR_{ext} = 2 \Omega + 8 \Omega Rext=10ΩR_{ext} = 10 \Omega

  4. Use the EMF equation to find the internal resistance rr.

    Given EMF, E=24 VE = 24 \text{ V}.
    The total current from the battery is given by:

    E=Itotal(Rext+r)E = I_{total}(R_{ext} + r) 24 V=1.2 A(10Ω+r)24 \text{ V} = 1.2 \text{ A}(10 \Omega + r)

    Divide both sides by 1.2 A1.2 \text{ A}:

    24 V1.2 A=10Ω+r\frac{24 \text{ V}}{1.2 \text{ A}} = 10 \Omega + r 20Ω=10Ω+r20 \Omega = 10 \Omega + r r=20Ω−10Ωr = 20 \Omega - 10 \Omega r=10Ωr = 10 \Omega

    The internal resistance is 10Ω\boxed{\text{}10 \Omega}.

Part b(iii)

12 V\text{}12 \text{ V}

The terminal potential difference of the battery.

  1. Use the formula for terminal potential difference.

    The terminal potential difference (VTV_T) is given by:

    VT=E−ItotalrV_T = E - I_{total}r

    Given:
    EMF, E=24 VE = 24 \text{ V}
    Total current, Itotal=1.2 AI_{total} = 1.2 \text{ A} (from part ii))
    Internal resistance, r=10Ωr = 10 \Omega (from part ii))

  2. Calculate VTV_T.

    VT=24 V−(1.2 A)(10Ω)V_T = 24 \text{ V} - (1.2 \text{ A})(10 \Omega) VT=24 V−12 VV_T = 24 \text{ V} - 12 \text{ V} VT=12 VV_T = 12 \text{ V}

    The terminal potential difference of the battery is 12 V\boxed{\text{}12 \text{ V}}.

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