The electric field (E) due to a point charge (Q) is given by E = kQ/r^2, where k is Coulomb's constant.

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.

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

Inserting a dielectric material increases the capacitance of the capacitor by a factor equal to the dielectric constant of the material.

Inserting a dielectric material between the plates of a capacitor increases its capacitance due to the material's ability to reduce the electric field.

The capacitance increases when a dielectric material is introduced, as it reduces the electric field between the plates.

When a capacitor is disconnected from a battery and the plates are moved further apart, the electric potential energy increases because the capacitance decreases while the charge remains constant.

Using the formula C = Q/V, we have C = 6μC / 3V = 2μF.

The capacitance of a spherical capacitor is given by C = 4πε₀ab/(b-a).

The dielectric constant (κ) of a vacuum is defined as 1, which serves as the reference for other materials.

When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances, thus it increases.

Introducing a dielectric material between the plates of a capacitor increases its capacitance by a factor equal to the dielectric constant (κ) of the material.

Increasing the area A of the plates increases the capacitance, as C = εA/d.

Increasing the distance between the plates of a capacitor decreases its capacitance, as capacitance is inversely proportional to the distance.

Increasing the plate area A of a parallel plate capacitor increases its capacitance, as C = ε₀A/d.

Introducing a dielectric material increases the capacitance of a capacitor by reducing the electric field between the plates.

The electrostatic potential V at a distance r from a point charge Q is given by V = kQ/r, where k is Coulomb's constant.

The energy stored (U) is given by U = 1/2 CV². Thus, U = 1/2 * 4μF * (12V)² = 288μJ.

Using the formula U = 1/2 CV², we have U = 1/2 * 5μF * (10V)² = 0.5mJ.

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

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

The potential energy (U) stored in a capacitor is given by the formula U = 1/2 CV², where C is the capacitance and V is the voltage across the capacitor.

The relationship is given by Q = CV, where Q is the charge, C is the capacitance, and V is the voltage across the capacitor.

In a uniform electric field, the electric field E is related to the potential difference V and the distance d by the equation E = V/d.

Glass is commonly used as a dielectric material in capacitors due to its high dielectric constant and insulating properties.

For capacitors in parallel, the total capacitance is the sum of the individual capacitances: C_total = C₁ + C₂ + ...