The charge stored in a capacitor is given by Q = C * V = 10μF * 5V = 50μC.

The charge stored in a capacitor is given by Q = C * V = 10μF * 5V = 50μC.

Charge (Q) is given by Q = C * V = 5μF * 10V = 0.05C.

Using Ohm's Law, V = I * R = 5A * 10Ω = 50V.

Using Ohm's Law, V = I * R = 2A * 10Ω = 20V.

According to Ohm's Law, if voltage is doubled and resistance is constant, current will also double.

Total resistance = 4Ω + 8Ω = 12Ω. Current I = V / R = 12V / 12Ω = 1A. Voltage drop across 8Ω = I * R = 1A * 8Ω = 8V.

Using Ohm's Law, I = V / R = 24V / 6Ω = 4A.

Total resistance R_total = R1 + R2 = 4Ω + 8Ω = 12Ω. Using Ohm's Law, I = V / R_total = 12V / 12Ω = 1A.

Using Coulomb's law, E = k * |Q| / r^2 = (8.99 x 10^9 N m²/C²) * (10 x 10^-6 C) / (2 m)^2 = 2247.5 N/C, approximately 2250 N/C.

For capacitors in series, 1/C_eq = 1/C1 + 1/C2 => C_eq = 2.4μF.

In parallel, the equivalent resistance is given by 1/R_eq = 1/R1 + 1/R2, so 1/R_eq = 1/6 + 1/3 = 1/2, thus R_eq = 2Ω.

In series, the equivalent resistance is R_eq = R1 + R2 = 6Ω + 3Ω = 9Ω.

Power (P) is given by P = I^2 * R = (3A)^2 * 10Ω = 90W.

Ohm's Law states that current is equal to voltage divided by resistance, I = V / R.

For capacitors in series, 1/C_eq = 1/C1 + 1/C2, so 1/C_eq = 1/4 + 1/6 = 5/12, thus C_eq = 2.4μF.

First, find the equivalent resistance: 1/R_eq = 1/4 + 1/6 => R_eq = 2.4Ω. Then, I = V / R_eq = 12V / 2.4Ω = 5A.

The unit of electric potential difference is the Volt (V).