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

When the plates of a disconnected capacitor are moved apart, the capacitance decreases, leading to an increase in potential difference.

Capacitance C is given by C = Q/V. Here, C = 24μC / 12V = 2μF.

The capacitance C is defined as C = Q/V.

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.

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

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

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

Capacitance C is inversely proportional to the distance d between the plates. If d is halved, C doubles, so the new capacitance is 10μF.

When the voltage is increased to 10 V, the charge on the capacitor increases because charge (Q) is directly proportional to voltage (V) for a given capacitance.

Capacitance C is defined as C = Q/V. If C is doubled and Q remains constant, then V must halve to maintain the equation.

The capacitance C of a parallel plate capacitor is given by C = εA/d. If the distance d is doubled, the new capacitance will be C/2.

If the distance d is halved, the capacitance C = εA/d will double.

According to Coulomb's law, the force F is inversely proportional to the square of the distance (F ∝ 1/r^2). If r is tripled, F becomes 1/9 of its original value.

The potential difference (V) between the plates is given by V = Ed, where E is the electric field and d is the distance between the plates.

The energy stored in a capacitor is given by U = 1/2 CV^2. If V is halved, the energy becomes U' = 1/2 C(V/2)^2 = 1/8 CV^2, which is halved.

The charge (Q) stored in a capacitor is directly proportional to the potential difference (V) across it, given by Q = CV.

Work done (W) = Charge (Q) × Potential difference (V) = 2 C × 10 V = 20 J.

Increasing the plate area while keeping the distance constant increases the capacitance, as capacitance is directly proportional to the area of the plates.

Capacitance (C) is directly proportional to the area of the plates (A) and the dielectric constant (ε) and inversely proportional to the distance (d) between the plates.

The capacitance (C) of a parallel plate capacitor is given by the formula C = ε₀A/d, where A is the area of the plates and d is the separation between them.

The electric field (E) between the plates of a parallel plate capacitor is given by the formula E = V/d, where V is the voltage and d is the separation between the plates.

The electric field (E) between the plates of a parallel plate capacitor is given by the expression E = V/d, where V is the voltage and d is the distance between the plates.

The electric potential (V) due to a point charge (Q) at a distance (r) is given by V = kQ/r, where k is Coulomb's constant.

The energy stored in a capacitor is given by U = 1/2 CV^2, which shows it is proportional to both charge and voltage.

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

The potential energy (U) stored in a capacitor is given by U = 1/2 CV^2, which shows it is directly proportional to both charge and voltage.

The potential energy (U) stored in a capacitor is given by U = 1/2 CV^2.