Step 1: Identify given values and the principle for equilibrium. The oil drop is stationary, meaning the upward electric force ($F_e$) balances the downward gravitational force ($F_g$). Given: Mass, $m = 5 \times 10^{-15} \text{ kg}$ Acceleration due to gravity, $g = 10 \text{ N/kg}$ Distance between plates, $d = 25 \text{ mm} = 0.025 \text{ m}$ Potential difference, $V = 1000 \text{ V}$ Step 2: Calculate the gravitational force. $$F_g = mg$$ $$F_g = (5 \times 10^{-15} \text{ kg})(10 \text{ N/kg})$$ $$F_g = 5 \times 10^{-14} \text{ N}$$ Step 3: Calculate the electric field between the plates. $$E = \frac{V}{d}$$ $$E = \frac{1000 \text{ V}}{0.025 \text{ m}}$$ $$E = 40000 \text{ V/m}$$ Step 4: Equate electric force and gravitational force to find the charge. $$F_e = F_g$$ $$qE = mg$$ $$q = \frac{mg}{E}$$ $$q = \frac{5 \times 10^{-14} \text{ N}}{40000 \text{ V/m}}$$ $$q = \frac{5 \times 10^{-14}}{4 \times 10^4} \text{ C}$$ $$q = 1.25 \times 10^{-18} \text{ C}$$ The charge on the drop is $\boxed{\text{}1.25 \times 10^{-18} \text{ C}}$. a) The capacitance of $C_1$ and $C_2$ in series. Step 1: Identify given values and the formula for the time constant. Given: Resistance, $R = 80 \text{ k}\Omega = 80 \times 10^3 \Omega$ Time constant, $\tau = 4.8 \text{ s}$ The time constant for an RC circuit is given by $\tau = RC_{eq}$, where $C_{eq}$ is the equivalent capacitance. Step 2: Calculate the equivalent capacitance $C_{eq}$. $$\tau = RC_{eq}$$ $$C_{eq} = \frac{\tau}{R}$$ $$C_{eq} = \frac{4.8 \text{ s}}{80 \times 10^3 \Omega}$$ $$C_{eq} = 6 \times 10^{-5} \text{ F}$$ $$C_{eq} = 60 \times 10^{-6} \text{ F} = 60 \mu\text{F}$$ The capacitance of $C_1$ and $C_2$ in series is $\boxed{\text{}60 \mu\text{F}}$. b) The capacitance of $C_1$ if $C_2$ has a capacitance of $100 \text{ mF}$. Step 1: Identify given values and the formula for capacitors in series. Given: Equivalent capacitance, $C_{eq} = 6 \times 10^{-5} \text{ F}$ (from part a)) Capacitance of $C_2 = 100 \text{ mF} = 100 \times 10^{-3} \text{ F} = 0.1 \text{ F}$ For capacitors in series, the equivalent capacitance is given by: $$\frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2}$$ Step 2: Solve for $C_1$. $$\frac{1}{C_1} = \frac{1}{C_{eq}} - \frac{1}{C_2}$$ $$\frac{1}{C_1} = \frac{1}{6 \times 10^{-5} \text{ F}} - \frac{1}{0.1 \text{ F}}$$ $$\frac{1}{C_1} = 16666.666... \text{ F}^{-1} - 10 \text{ F}^{-1}$$ $$\frac{1}{C_1} = 16656.666... \text{ F}^{-1}$$ $$C_1 = \frac{1}{16656.666...} \text{ F}$$ $$C_1 \approx 6.003 \times 10^{-5} \text{ F}$$ $$C_1 \approx 60.03 \mu\text{F}$$ The capacitance of $C_1$ is $\boxed{\text{}60.03 \mu\text{F}}$. i) The current in the $6 \Omega$ resistor. Step 1: Identify given values and the circuit configuration. Given: Current in $3 \Omega$ resistor, $I_{3\Omega} = 0.8 \text{ A}$ The $3 \Omega$ and $6 \Omega$ resistors are connected in parallel. In a parallel circuit, the voltage across each branch is the same. Step 2: Calculate the voltage across the $3 \Omega$ resistor. Using Ohm's Law, $V = IR$: $$V_{3\Omega} = I_{3\Omega} \times R_{3\Omega}$$ $$V_{3\Omega} = 0.8 \text{ A} \times 3 \Omega$$ $$V_{3\Omega} = 2.4 \text{ V}$$ Step 3: Calculate the current in the $6 \Omega$ resistor. Since the resistors are in parallel, $V_{6\Omega} = V_{3\Omega} = 2.4 \text{ V}$. $$I_{6\Omega} = \frac{V_{6\Omega}}{R_{6\Omega}}$$ $$I_{6\Omega} = \frac{2.4 \text{ V}}{6 \Omega}$$ $$I_{6\Omega} = 0.4 \text{ A}$$ The current in the $6 \Omega$ resistor is $\boxed{\text{}0.4 \text{ A}}$. ii) The internal resistance, $r$. Step 1: Calculate the total current flowing from the battery. The total current ($I_{total}$) is the sum of the currents in the parallel branches. $$I_{total} = I_{3\Omega} + I_{6\Omega}$$ $$I_{total} = 0.8 \text{ A} + 0.4 \text{ A}$$ $$I_{total} = 1.2 \text{ A}$$ Step 2: Calculate the equivalent resistance of the parallel combination. For parallel resistors: $$\frac{1}{R_p} = \frac{1}{R_{3\Omega}} + \frac{1}{R_{6\Omega}}$$ $$\frac{1}{R_p} = \frac{1}{3 \Omega} + \frac{1}{6 \Omega}$$ $$\frac{1}{R_p} = \frac{2}{6 \Omega} + \frac{1}{6 \Omega}$$ $$\frac{1}{R_p} = \frac{3}{6 \Omega}$$ $$R_p = \frac{6}{3} \Omega = 2 \Omega$$ Step 3: Calculate the total external resistance. The parallel combination ($R_p$) is in series with the $8 \Omega$ resistor. $$R_{ext} = R_p + 8 \Omega$$ $$R_{ext} = 2 \Omega + 8 \Omega$$ $$R_{ext} = 10 \Omega$$ Step 4: Use the EMF equation to find the internal resistance $r$. Given EMF, $E = 24 \text{ V}$. The total current from the battery is given by: $$E = I_{total}(R_{ext} + r)$$ $$24 \text{ V} = 1.2 \text{ A}(10 \Omega + r)$$ Divide both sides by $1.2 \text{ A}$: $$\frac{24 \text{ V}}{1.2 \text{ A}} = 10 \Omega + r$$ $$20 \Omega = 10 \Omega + r$$ $$r = 20 \Omega - 10 \Omega$$ $$r = 10 \Omega$$ The internal resistance is $\boxed{\text{}10 \Omega}$. iii) The terminal potential difference of the battery. Step 1: Use the formula for terminal potential difference. The terminal potential difference ($V_T$) is given by: $$V_T = E - I_{total}r$$ Given: EMF, $E = 24 \text{ V}$ Total current, $I_{total} = 1.2 \text{ A}$ (from part ii)) Internal resistance, $r = 10 \Omega$ (from part ii)) Step 2: Calculate $V_T$. $$V_T = 24 \text{ V} - (1.2 \text{ A})(10 \Omega)$$ $$V_T = 24 \text{ V} - 12 \text{ V}$$ $$V_T = 12 \text{ V}$$ The terminal potential difference of the battery is $\boxed{\text{}12 \text{ V}}$.