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Electrostatic Potential and Capacitance - Effect of Dielectric on Capacitance

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 by an external electric field.

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When a dielectric slab is inserted between the plates of a capacitor, the induced charges on the surface of the dielectric create an internal electric field EpE_p opposite to the external field E0E_0.

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The net electric field inside the dielectric is reduced to E=E0−Ep=E0KE = E_0 - E_p = \frac{E_0}{K}, where KK is the dielectric constant (relative permittivity).

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The potential difference between the plates reduces by the same factor if the capacitor is isolated: V=V0KV = \frac{V_0}{K}.

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The capacitance of a parallel plate capacitor increases by a factor of KK: C=KC0=Kϵ0AdC = K C_0 = \frac{K \epsilon_0 A}{d}.

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If a dielectric slab of thickness tt (where t<dt < d) is inserted, the capacitance becomes C=ϵ0Ad−t+tKC = \frac{\epsilon_0 A}{d - t + \frac{t}{K}}.

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Effect of Dielectric when the battery remains connected: Potential VV remains constant, Capacitance CC increases to KC0KC_0, Charge QQ increases to KQ0KQ_0, and Energy UU increases to KU0KU_0.

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Effect of Dielectric when the battery is disconnected: Charge QQ remains constant, Capacitance CC increases to KC0KC_0, Potential VV decreases to V0K\frac{V_0}{K}, and Energy UU decreases to U0K\frac{U_0}{K}.

📐Formulae

K=ϵϵ0K = \frac{\epsilon}{\epsilon_0}

C=Kϵ0AdC = \frac{K \epsilon_0 A}{d}

V=V0KV = \frac{V_0}{K}

E=E0KE = \frac{E_0}{K}

Qp=Q(1−1K)Q_{p} = Q \left( 1 - \frac{1}{K} \right)

C=ϵ0Ad−t(1−1K)C = \frac{\epsilon_0 A}{d - t(1 - \frac{1}{K})}

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:

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

Explanation:

Capacitance is directly proportional to the dielectric constant and inversely proportional to the distance between the plates. Halving the distance doubles the capacitance, and adding a dielectric of K=6K=6 increases it by 66 times, resulting in a total increase of 1212 times.

Problem 2:

A 12 pF12\text{ pF} capacitor is connected to a 50 V50\text{ V} battery. The battery is then disconnected, and a mica sheet (K=6K = 6) is inserted between the plates. Calculate the change in energy stored in the capacitor.

Solution:

Initial Energy Ui=12C0V02=12×12×10−12×502=15000×10−12=1.5×10−8 JU_i = \frac{1}{2} C_0 V_0^2 = \frac{1}{2} \times 12 \times 10^{-12} \times 50^2 = 15000 \times 10^{-12} = 1.5 \times 10^{-8}\text{ J}. Since the battery is disconnected, charge QQ remains constant. The new energy Uf=UiKU_f = \frac{U_i}{K}. Uf=1.5×10−86=0.25×10−8 JU_f = \frac{1.5 \times 10^{-8}}{6} = 0.25 \times 10^{-8}\text{ J} The change in energy ΔU=Ui−Uf\Delta U = U_i - U_f (energy is lost): 1.50×10−8−0.25×10−81.25×10−8\begin{array}{r} 1.50 \times 10^{-8} \\ - 0.25 \times 10^{-8} \\ \hline 1.25 \times 10^{-8} \end{array} Thus, ΔU=1.25×10−8 J\Delta U = 1.25 \times 10^{-8}\text{ J}.

Explanation:

When the battery is disconnected, the charge is trapped. Inserting a dielectric increases capacitance, which lowers the potential energy (U=Q2/2CU = Q^2/2C) as work is done by the field to pull the dielectric slab inside.