1.;A capacitor has a capacitance of 8.0 microFarad with air between its plates. Determine its capacitance when a dielectric with dielectric constant 6.0 is placed between its plates. 2. What is the charge on a 300 pF capacitor when it is charged to a voltage of 1.0 kV? 3. A certain parallel-plate capacitor consists of two plates, each with area 200 cm² , separated by a 0.40 cm air gap. (a) Compute its capacitance. (b) If the capacitor is connected across a 500 V source, find the charge on it, the energy stored in it, and the value of E between the plates. (c) If a liquid with K = 2.60 is poured between the plates so as to fill the air gap, how much additional charge will flow onto the capacitor from the 500 V source?

Question

1.;A capacitor has a capacitance of 8.0 microFarad with air between its plates. Determine its capacitance when a dielectric with dielectric constant 6.0 is placed between its plates. 2. What is the charge on a 300 pF capacitor when it is charged to a voltage of 1.0 kV? 3. A certain parallel-plate capacitor consists of two plates, each with area 200 cm² , separated by a 0.40 cm air gap. (a) Compute its capacitance. (b) If the capacitor is connected across a 500 V source, find the charge on it, the energy stored in it, and the value of E between the plates. (c) If a liquid with K = 2.60 is poured between the plates so as to fill the air gap, how much additional charge will flow onto the capacitor from the 500 V source?

Asked on April 13, 2026Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

Accepted Answer · 2026-04-13T14:35:26.138Z

Here are the solutions to your capacitor problems. 1. A capacitor has a capacitance of 8.0 microFarad with air between its plates. Determine its capacitance when a dielectric with dielectric constant 6.0 is placed between its plates. Step 1: Identify the given values. Initial capacitance with air, $C_0 = 8.0 \text{ µF}$. Dielectric constant, $K = 6.0$. Step 2: Use the formula for capacitance with a dielectric. The capacitance $C$ with a dielectric is given by $C = K C_0$. Step 3: Substitute the values and calculate the new capacitance. $$ C = (6.0) \times (8.0 \text{ µF}) $$ $$ C = 48.0 \text{ µF} $$ The capacitance with the dielectric is $\boxed{48.0 \text{ µF}}$. 2. What is the charge on a 300 pF capacitor when it is charged to a voltage of 1.0 kV? Step 1: Identify the given values and convert units to SI units. Capacitance, $C = 300 \text{ pF} = 300 \times 10^{-12} \text{ F}$. Voltage, $V = 1.0 \text{ kV} = 1.0 \times 10^3 \text{ V}$. Step 2: Use the formula relating charge, capacitance, and voltage. The charge $Q$ on a capacitor is given by $Q = CV$. Step 3: Substitute the values and calculate the charge. $$ Q = (300 \times 10^{-12} \text{ F}) \times (1.0 \times 10^3 \text{ V}) $$ $$ Q = 300 \times 10^{-9} \text{ C} $$ $$ Q = 3.0 \times 10^{-7} \text{ C} $$ The charge on the capacitor is $\boxed{3.0 \times 10^{-7} \text{ C}}$. 3. A certain parallel-plate capacitor consists of two plates, each with area 200 cm² , separated by a 0.40 cm air gap. (a) Compute its capacitance. (b) If the capacitor is connected across a 500 V source, find the charge on it, the energy stored in it, and the value of E between the plates. (c) If a liquid with K = 2.60 is poured between the plates so as to fill the air gap, how much additional charge will flow onto the capacitor from the 500 V source? Given values: Area, $A = 200 \text{ cm}^2 = 200 \times (10^{-2} \text{ m})^2 = 200 \times 10^{-4} \text{ m}^2 = 0.0200 \text{ m}^2$. Separation, $d = 0.40 \text{ cm} = 0.40 \times 10^{-2} \text{ m} = 0.0040 \text{ m}$. Permittivity of free space, $\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$. Voltage, $V = 500 \text{ V}$. Dielectric constant, $K = 2.60$. (a) Compute its capacitance. Step 1: Use the formula for the capacitance of a parallel-plate capacitor with air. $$ C_0 = \frac{\epsilon_0 A}{d} $$ Step 2: Substitute the values and calculate the capacitance. $$ C_0 = \frac{(8.85 \times 10^{-12} \text{ F/m}) \times (0.0200 \text{ m}^2)}{0.0040 \text{ m}} $$ $$ C_0 = \frac{1.77 \times 10^{-13} \text{ F}\cdot\text{m}}{0.0040 \text{ m}} $$ $$ C_0 = 4.425 \times 10^{-11} \text{ F} $$ $$ C_0 = 44.25 \text{ pF} $$ The capacitance is $\boxed{4.43 \times 10^{-11} \text{ F}}$ (or $44.3 \text{ pF}$). (b) If the capacitor is connected across a 500 V source, find the charge on it, the energy stored in it, and the value of E between the plates. Step 1: Calculate the charge $Q_0$ on the capacitor. $$ Q_0 = C_0 V $$ $$ Q_0 = (4.425 \times 10^{-11} \text{ F}) \times (500 \text{ V}) $$ $$ Q_0 = 2.2125 \times 10^{-8} \text{ C} $$ The charge on the capacitor is $\boxed{2.21 \times 10^{-8} \text{ C}}$. Step 2: Calculate the energy stored $U_0$ in the capacitor. $$ U_0 = \frac{1}{2} C_0 V^2 $$ $$ U_0 = \frac{1}{2} (4.425 \times 10^{-11} \text{ F}) \times (500 \text{ V})^2 $$ $$ U_0 = \frac{1}{2} (4.425 \times 10^{-11} \text{ F}) \times (250000 \text{ V}^2) $$ $$ U_0 = 5.53125 \times 10^{-6} \text{ J} $$ The energy stored in the capacitor is $\boxed{5.53 \times 10^{-6} \text{ J}}$. Step 3: Calculate the electric field $E_0$ between the plates. $$ E_0 = \frac{V}{d} $$ $$ E_0 = \frac{500 \text{ V}}{0.0040 \text{ m}} $$ $$ E_0 = 125000 \text{ V/m} $$ The electric field between the plates is $\boxed{1.25 \times 10^5 \text{ V/m}}$. (c) If a liquid with K = 2.60 is poured between the plates so as to fill the air gap, how much additional charge will flow onto the capacitor from the 500 V source? Step 1: Calculate the new capacitance $C$ with the dielectric. $$ C = K C_0 $$ $$ C = (2.60) \times (4.425 \times 10^{-11} \text{ F}) $$ $$ C = 1.1505 \times 10^{-10} \text{ F} $$ Step 2: Calculate the new charge $Q$ on the capacitor with the dielectric. The voltage source remains $V = 500 \text{ V}$. $$ Q = C V $$ $$ Q = (1.1505 \times 10^{-10} \text{ F}) \times (500 \text{ V}) $$ $$ Q = 5.7525 \times 10^{-8} \text{ C} $$ Step 3: Calculate the additional charge $\Delta Q$ that flows onto the capacitor. $$ \Delta Q = Q - Q_0 $$ $$ \Delta Q = (5.7525 \times 10^{-8} \text{ C}) - (2.2125 \times 10^{-8} \text{ C}) $$ $$ \Delta Q = 3.54 \times 10^{-8} \text{ C} $$ The additional charge that will flow onto the capacitor is $\boxed{3.54 \times 10^{-8} \text{ C}}$. What's next?

Accepted Answer · 2026-04-13T14:35:26.138Z

Here are the solutions to your capacitor problems. 1. A capacitor has a capacitance of 8.0 microFarad with air between its plates. Determine its capacitance when a dielectric with dielectric constant 6.0 is placed between its plates. Step 1: Identify the given values. Initial capacitance with air, $C_0 = 8.0 \text{ µF}$. Dielectric constant, $K = 6.0$. Step 2: Use the formula for capacitance with a dielectric. The capacitance $C$ with a dielectric is given by $C = K C_0$. Step 3: Substitute the values and calculate the new capacitance. $$ C = (6.0) \times (8.0 \text{ µF}) $$ $$ C = 48.0 \text{ µF} $$ The capacitance with the dielectric is $\boxed{48.0 \text{ µF}}$. 2. What is the charge on a 300 pF capacitor when it is charged to a voltage of 1.0 kV? Step 1: Identify the given values and convert units to SI units. Capacitance, $C = 300 \text{ pF} = 300 \times 10^{-12} \text{ F}$. Voltage, $V = 1.0 \text{ kV} = 1.0 \times 10^3 \text{ V}$. Step 2: Use the formula relating charge, capacitance, and voltage. The charge $Q$ on a capacitor is given by $Q = CV$. Step 3: Substitute the values and calculate the charge. $$ Q = (300 \times 10^{-12} \text{ F}) \times (1.0 \times 10^3 \text{ V}) $$ $$ Q = 300 \times 10^{-9} \text{ C} $$ $$ Q = 3.0 \times 10^{-7} \text{ C} $$ The charge on the capacitor is $\boxed{3.0 \times 10^{-7} \text{ C}}$. 3. A certain parallel-plate capacitor consists of two plates, each with area 200 cm² , separated by a 0.40 cm air gap. (a) Compute its capacitance. (b) If the capacitor is connected across a 500 V source, find the charge on it, the energy stored in it, and the value of E between the plates. (c) If a liquid with K = 2.60 is poured between the plates so as to fill the air gap, how much additional charge will flow onto the capacitor from the 500 V source? Given values: Area, $A = 200 \text{ cm}^2 = 200 \times (10^{-2} \text{ m})^2 = 200 \times 10^{-4} \text{ m}^2 = 0.0200 \text{ m}^2$. Separation, $d = 0.40 \text{ cm} = 0.40 \times 10^{-2} \text{ m} = 0.0040 \text{ m}$. Permittivity of free space, $\epsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$. Voltage, $V = 500 \text{ V}$. Dielectric constant, $K = 2.60$. (a) Compute its capacitance. Step 1: Use the formula for the capacitance of a parallel-plate capacitor with air. $$ C_0 = \frac{\epsilon_0 A}{d} $$ Step 2: Substitute the values and calculate the capacitance. $$ C_0 = \frac{(8.85 \times 10^{-12} \text{ F/m}) \times (0.0200 \text{ m}^2)}{0.0040 \text{ m}} $$ $$ C_0 = \frac{1.77 \times 10^{-13} \text{ F}\cdot\text{m}}{0.0040 \text{ m}} $$ $$ C_0 = 4.425 \times 10^{-11} \text{ F} $$ $$ C_0 = 44.25 \text{ pF} $$ The capacitance is $\boxed{4.43 \times 10^{-11} \text{ F}}$ (or $44.3 \text{ pF}$). (b) If the capacitor is connected across a 500 V source, find the charge on it, the energy stored in it, and the value of E between the plates. Step 1: Calculate the charge $Q_0$ on the capacitor. $$ Q_0 = C_0 V $$ $$ Q_0 = (4.425 \times 10^{-11} \text{ F}) \times (500 \text{ V}) $$ $$ Q_0 = 2.2125 \times 10^{-8} \text{ C} $$ The charge on the capacitor is $\boxed{2.21 \times 10^{-8} \text{ C}}$. Step 2: Calculate the energy stored $U_0$ in the capacitor. $$ U_0 = \frac{1}{2} C_0 V^2 $$ $$ U_0 = \frac{1}{2} (4.425 \times 10^{-11} \text{ F}) \times (500 \text{ V})^2 $$ $$ U_0 = \frac{1}{2} (4.425 \times 10^{-11} \text{ F}) \times (250000 \text{ V}^2) $$ $$ U_0 = 5.53125 \times 10^{-6} \text{ J} $$ The energy stored in the capacitor is $\boxed{5.53 \times 10^{-6} \text{ J}}$. Step 3: Calculate the electric field $E_0$ between the plates. $$ E_0 = \frac{V}{d} $$ $$ E_0 = \frac{500 \text{ V}}{0.0040 \text{ m}} $$ $$ E_0 = 125000 \text{ V/m} $$ The electric field between the plates is $\boxed{1.25 \times 10^5 \text{ V/m}}$. (c) If a liquid with K = 2.60 is poured between the plates so as to fill the air gap, how much additional charge will flow onto the capacitor from the 500 V source? Step 1: Calculate the new capacitance $C$ with the dielectric. $$ C = K C_0 $$ $$ C = (2.60) \times (4.425 \times 10^{-11} \text{ F}) $$ $$ C = 1.1505 \times 10^{-10} \text{ F} $$ Step 2: Calculate the new charge $Q$ on the capacitor with the dielectric. The voltage source remains $V = 500 \text{ V}$. $$ Q = C V $$ $$ Q = (1.1505 \times 10^{-10} \text{ F}) \times (500 \text{ V}) $$ $$ Q = 5.7525 \times 10^{-8} \text{ C} $$ Step 3: Calculate the additional charge $\Delta Q$ that flows onto the capacitor. $$ \Delta Q = Q - Q_0 $$ $$ \Delta Q = (5.7525 \times 10^{-8} \text{ C}) - (2.2125 \times 10^{-8} \text{ C}) $$ $$ \Delta Q = 3.54 \times 10^{-8} \text{ C} $$ The additional charge that will flow onto the capacitor is $\boxed{3.54 \times 10^{-8} \text{ C}}$. What's next?

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