Charge (Q) is given by Q = CV. Here, Q = 3 µF * 15 V = 45 µC.

If the voltage is doubled, the charge on the capacitor also doubles, since Q = C * V.

When two identical capacitors are connected in parallel, the charge is shared equally. The new voltage is halved: 20 V / 2 = 10 V.

In an AC circuit, the current through a capacitor leads the voltage by 90 degrees.

The capacitance increases by a factor equal to the dielectric constant. Thus, it quadruples.

Charge (Q) is given by Q = C * V. Here, Q = 3 µF * 6 V = 18 µC.

Total resistance in series is R_total = R1 + R2 = 10 Ω + 20 Ω = 30 Ω.

Total resistance R = 4 Ω + 6 Ω = 10 Ω. Current I = V/R = 12 V / 10 Ω = 1.2 A. Voltage across 4 Ω = I * R = 1.2 A * 4 Ω = 4.8 V.

Total capacitance in series is given by 1/C_total = 1/C1 + 1/C2. Thus, 1/C_total = 1/3 + 1/6 = 1/2, so C_total = 2 µF.

Capacitance is directly proportional to the plate area. Doubling the area doubles the capacitance.

Capacitance (C) is given by C = Q/V. Here, C = 5 µC / 10 V = 0.5 µF.

Adding a dielectric increases the capacitance of the capacitor.

A capacitor stores energy when connected in parallel with a resistor, affecting the circuit's behavior over time.

Capacitance decreases as the distance between the plates increases.

Capacitance (C) is inversely proportional to the distance (d) between the plates. Increasing d decreases C.

Electric potential energy (U) is given by U = 1/2 CV^2. Here, U = 1/2 * 2 µF * (12 V)^2 = 0.144 mJ.

Electric potential energy (U) is given by U = 1/2 C V^2. Here, U = 1/2 * 2 µF * (12 V)^2 = 0.144 mJ.

Energy (U) is given by U = 1/2 CV^2. Here, U = 1/2 * 5 µF * (10 V)^2 = 0.5 mJ.

For capacitors in series, 1/C_eq = 1/C1 + 1/C2. Thus, 1/C_eq = 1/2 + 1/3 = 5/6. Therefore, C_eq = 6/5 = 1.2 µF.

The electric field between two parallel plates is given by E = V/d.

Using C = Q/V, we find V = Q/C = 10 µC / 5 µF = 2 V.

Using the formula V = Q/C, V = 20 µC / 5 µF = 4 V.