;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

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 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) × (8.0 µF) C = 48.0 µF The capacitance with the dielectric is 48.0 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 pF = 300 × 10^-12 F. Voltage, V = 1.0 kV = 1.0 × 10^3 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 × 10^-12 F) × (1.0 × 10^3 V) Q = 300 × 10^-9 C Q = 3.0 × 10^-7 C The charge on the capacitor is 3.0 × 10^-7 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 cm^2 = 200 × (10^-2 m)^2 = 200 × 10^-4 m^2 = 0.0200 m^2. Separation, d = 0.40 cm = 0.40 × 10^-2 m = 0.0040 m. Permittivity of free space, _0 = 8.85 × 10^-12 F/m. Voltage, V = 500 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 = (_0 A)/(d) Step 2: Substitute the values and calculate the capacitance. C_0 = (8.85 × 10^-12 F/m) × (0.0200 m^2)0.0040 m C_0 = 1.77 × 10^-13 F·m0.0040 m C_0 = 4.425 × 10^-11 F C_0 = 44.25 pF The capacitance is 4.43 × 10^-11 F (or 44.3 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 × 10^-11 F) × (500 V) Q_0 = 2.2125 × 10^-8 C The charge on the capacitor is 2.21 × 10^-8 C. Step 2: Calculate the energy stored U_0 in the capacitor. U_0 = (1)/(2) C_0 V^2 U_0 = (1)/(2) (4.425 × 10^-11 F) × (500 V)^2 U_0 = (1)/(2) (4.425 × 10^-11 F) × (250000 V^2) U_0 = 5.53125 × 10^-6 J The energy stored in the capacitor is 5.53 × 10^-6 J. Step 3: Calculate the electric field E_0 between the plates. E_0 = (V)/(d) E_0 = 500 V0.0040 m E_0 = 125000 V/m The electric field between the plates is 1.25 × 10^5 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) × (4.425 × 10^-11 F) C = 1.1505 × 10^-10 F Step 2: Calculate the new charge Q on the capacitor with the dielectric. The voltage source remains V = 500 V. Q = C V Q = (1.1505 × 10^-10 F) × (500 V) Q = 5.7525 × 10^-8 C Step 3: Calculate the additional charge Q that flows onto the capacitor. Q = Q - Q_0 Q = (5.7525 × 10^-8 C) - (2.2125 × 10^-8 C) Q = 3.54 × 10^-8 C The additional charge that will flow onto the capacitor is 3.54 × 10^-8 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 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) × (8.0 µF) C = 48.0 µF The capacitance with the dielectric is 48.0 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 pF = 300 × 10^-12 F. Voltage, V = 1.0 kV = 1.0 × 10^3 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 × 10^-12 F) × (1.0 × 10^3 V) Q = 300 × 10^-9 C Q = 3.0 × 10^-7 C The charge on the capacitor is 3.0 × 10^-7 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 cm^2 = 200 × (10^-2 m)^2 = 200 × 10^-4 m^2 = 0.0200 m^2. Separation, d = 0.40 cm = 0.40 × 10^-2 m = 0.0040 m. Permittivity of free space, _0 = 8.85 × 10^-12 F/m. Voltage, V = 500 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 = (_0 A)/(d) Step 2: Substitute the values and calculate the capacitance. C_0 = (8.85 × 10^-12 F/m) × (0.0200 m^2)0.0040 m C_0 = 1.77 × 10^-13 F·m0.0040 m C_0 = 4.425 × 10^-11 F C_0 = 44.25 pF The capacitance is 4.43 × 10^-11 F (or 44.3 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 × 10^-11 F) × (500 V) Q_0 = 2.2125 × 10^-8 C The charge on the capacitor is 2.21 × 10^-8 C. Step 2: Calculate the energy stored U_0 in the capacitor. U_0 = (1)/(2) C_0 V^2 U_0 = (1)/(2) (4.425 × 10^-11 F) × (500 V)^2 U_0 = (1)/(2) (4.425 × 10^-11 F) × (250000 V^2) U_0 = 5.53125 × 10^-6 J The energy stored in the capacitor is 5.53 × 10^-6 J. Step 3: Calculate the electric field E_0 between the plates. E_0 = (V)/(d) E_0 = 500 V0.0040 m E_0 = 125000 V/m The electric field between the plates is 1.25 × 10^5 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) × (4.425 × 10^-11 F) C = 1.1505 × 10^-10 F Step 2: Calculate the new charge Q on the capacitor with the dielectric. The voltage source remains V = 500 V. Q = C V Q = (1.1505 × 10^-10 F) × (500 V) Q = 5.7525 × 10^-8 C Step 3: Calculate the additional charge Q that flows onto the capacitor. Q = Q - Q_0 Q = (5.7525 × 10^-8 C) - (2.2125 × 10^-8 C) Q = 3.54 × 10^-8 C The additional charge that will flow onto the capacitor is 3.54 × 10^-8 C. What's next?

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