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Electrostatic Potential and Capacitance - Dielectrics and Polarization

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Dielectrics are non-conducting substances that do not have free charge carriers but can be polarized in the presence of an external electric field E0E_0.

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Non-polar molecules (e.g., H2H_2, O2O_2, CO2CO_2) are those in which the centers of positive and negative charges coincide, resulting in zero intrinsic dipole moment.

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Polar molecules (e.g., H2OH_2O, HClHCl) have centers of positive and negative charges that are separated even in the absence of an electric field, giving them a permanent dipole moment.

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Polarization (PP) is defined as the dipole moment per unit volume. For linear isotropic dielectrics, P=χeϵ0EP = \chi_e \epsilon_0 E, where χe\chi_e is the electric susceptibility.

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When a dielectric slab is placed in an external field E0E_0, an internal electric field EpE_p is induced in the opposite direction. The net reduced field inside the dielectric is E=E0−EpE = E_0 - E_p.

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The Dielectric Constant (KK) is the ratio of the external electric field to the reduced electric field: K=E0EK = \frac{E_0}{E}. It is also known as the relative permittivity ϵr\epsilon_r.

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Inserting a dielectric material between the plates of a capacitor increases its capacitance by a factor of KK. If C0C_0 is the original capacitance, the new capacitance C=KC0C = K C_0.

📐Formulae

E=E0−Ep=E0KE = E_0 - E_p = \frac{E_0}{K}

P=χeϵ0EP = \chi_e \epsilon_0 E

K=1+χeK = 1 + \chi_e

σp=σ(1−1K)\sigma_p = \sigma \left(1 - \frac{1}{K}\right)

C=Kϵ0AdC = \frac{K \epsilon_0 A}{d}

U=12CV2U = \frac{1}{2} C V^2

💡Examples

Problem 1:

A parallel plate capacitor with air between the plates has a capacitance of 8 pF8\text{ pF}. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant K=6K = 6?

Solution:

C0=ϵ0Ad=8 pFC_0 = \frac{\epsilon_0 A}{d} = 8\text{ pF} When the distance is d′=d2d' = \frac{d}{2} and the dielectric is K=6K=6, the new capacitance is: C=Kϵ0Ad/2=2K(ϵ0Ad)C = \frac{K \epsilon_0 A}{d/2} = 2K \left(\frac{\epsilon_0 A}{d}\right) C=2×6×8 pF=96 pFC = 2 \times 6 \times 8\text{ pF} = 96\text{ pF}

Explanation:

Capacitance is directly proportional to the dielectric constant KK and inversely proportional to the distance dd. Reducing dd to half doubles the capacitance, and filling it with K=6K=6 increases it by six times.

Problem 2:

A dielectric slab of constant KK is introduced between the plates of a charged capacitor which is disconnected from the battery. What happens to the energy stored in the capacitor?

Solution:

Let the initial charge be QQ and capacitance be C0C_0. Initial energy U0=Q22C0U_0 = \frac{Q^2}{2C_0}. When the battery is disconnected, QQ remains constant. With the dielectric, C=KC0C = K C_0. New energy U=Q22(KC0)=U0KU = \frac{Q^2}{2(K C_0)} = \frac{U_0}{K}

Explanation:

Since the charge remains constant and the capacitance increases by factor KK, the energy stored in the capacitor decreases by a factor of 1/K1/K because work is done by the field to pull the dielectric into the capacitor.

Dielectrics and Polarization - Revision Notes & Key Formulas | CBSE Class 12 Physics