Energy stored U = 1/2 CV² = 1/2 * 4 x 10^-6 F * (12 V)² = 0.24 mJ.

Charge (Q) = Capacitance (C) × Voltage (V) = 4 µF × 12 V = 48 µC.

Capacitance (C) = Charge (Q) / Voltage (V) = 6 µC / 12 V = 0.5 µF.

When the distance is doubled, the potential difference across the capacitor also doubles, resulting in 24V.

Once disconnected, the charge remains constant. Doubling the area increases capacitance but does not change the potential difference since the charge is fixed.

The potential V across a capacitor is directly proportional to the charge. If the charge is doubled, the potential also doubles.

The voltage across the capacitor decreases exponentially over time when connected to a resistor due to discharge.

When the distance between the plates is doubled, the capacitance decreases, which causes the voltage to double since Q remains constant.

The time constant (τ) of an RC circuit is given by τ = RC.

When a capacitor is disconnected from the battery, the charge remains constant. Increasing the distance decreases capacitance but does not affect the charge.

When the distance is doubled, the capacitance decreases, leading to an increase in voltage since Q = CV is constant.

When a capacitor is disconnected from the battery, the charge remains constant. Increasing the distance decreases capacitance but does not change the charge.

The charge on a capacitor is given by Q = C * V. If the voltage is doubled, the charge also doubles, assuming capacitance remains constant.

Once disconnected from the battery, the charge on the capacitor remains constant as there is no path for charge to flow.

The energy stored in a capacitor is proportional to the square of the voltage (U = 1/2 CV²). If the voltage is halved, the energy becomes U/4.

The energy (U) stored in a capacitor is given by U = 1/2 CV^2, where C is the capacitance and V is the voltage.

Capacitive reactance XC = 1/(2πfC) = 1/(2π*50*10e-6) ≈ 318.3Ω.

Capacitive reactance Xc = 1/(ωC) where ω = 2πf. Calculate Xc.

Charge (Q) is given by Q = C * V. Thus, Q = 10 µF * 5 V = 50 µC.

Energy stored, U = 1/2 * C * V^2 = 1/2 * 10 × 10^-6 * (100)^2 = 0.05 J.

Energy stored in a capacitor is given by U = 1/2 CV² = 1/2 * 10 × 10^-6 F * (100 V)² = 0.05 J.

Charge (Q) = Capacitance (C) × Voltage (V) = 5µF × 10V = 50µC.

Energy stored U = 1/2 * C * V² = 1/2 * 5 × 10^-6 F * (10 V)² = 0.5 mJ.

When connected in parallel, the total charge is conserved. The final voltage across both capacitors is V/2.

When connected, charge redistributes between the two capacitors, resulting in a final voltage of V/2 across each.

If the charge is removed, the potential difference across the capacitor becomes 0 volts.

The energy stored in a capacitor is given by U = 1/2 CV². If the voltage is halved, the new energy becomes U' = 1/2 C(V/2)² = U/4.

The charge Q stored in a capacitor is given by Q = CV.

When the charged capacitor C is connected to an uncharged capacitor 2C, the final voltage is V_final = Q_total / C_eq = V/(1 + 1/2) = V/3.

The capacitive reactance is given by XC = 1/(2πfC).