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Electrostatics - Capacitors and Dielectrics

Grade 12ICSEPhysics

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

πŸ”‘Concepts

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Capacitance is defined as the ratio of the charge QQ on a conductor to its potential VV. The unit is the Farad (FF), where 1F=1C/V1 F = 1 C/V.

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A Parallel Plate Capacitor consists of two conducting plates of area AA separated by a distance dd. The capacitance depends on the geometry and the medium between the plates.

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Dielectrics are non-conducting substances that polarize in an external electric field. This polarization creates an internal electric field that opposes the external field, reducing the net electric field to E=E0/KE = E_0 / K.

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The Dielectric Constant (KK) is the ratio of the capacitance with the dielectric to the capacitance in a vacuum (K=C/C0K = C / C_0). It is also known as relative permittivity Ο΅r\epsilon_r.

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When capacitors are connected in series, the charge QQ on each capacitor is the same, but the total potential difference is the sum of individual potentials: V=V1+V2+…V = V_1 + V_2 + \dots.

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When capacitors are connected in parallel, the potential difference VV across each capacitor is the same, and the total charge is the sum of individual charges: Q=Q1+Q2+…Q = Q_1 + Q_2 + \dots.

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Energy is stored in the electric field between the plates. The energy density uu is the energy per unit volume, given by u=12Ο΅0E2u = \frac{1}{2} \epsilon_0 E^2.

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If a dielectric slab of thickness t<dt < d is inserted between plates of separation dd, the new capacitance is C=Ο΅0Adβˆ’t(1βˆ’1/K)C = \frac{\epsilon_0 A}{d - t(1 - 1/K)}.

πŸ“Formulae

C=QVC = \frac{Q}{V}

C=Ο΅0AdC = \frac{\epsilon_0 A}{d}

Cmedium=KCair=KΟ΅0AdC_{medium} = K C_{air} = \frac{K \epsilon_0 A}{d}

1Cs=1C1+1C2+1C3+…\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots

Cp=C1+C2+C3+…C_p = C_1 + C_2 + C_3 + \dots

U=12CV2=Q22C=12QVU = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV

u=12Ο΅0E2u = \frac{1}{2} \epsilon_0 E^2

V=Eβ‹…dV = E \cdot d

Οƒp=Οƒ(1βˆ’1K)\sigma_p = \sigma \left(1 - \frac{1}{K}\right)

πŸ’‘Examples

Problem 1:

A parallel plate capacitor has a plate area of 6Γ—10βˆ’3m26 \times 10^{-3} m^2 and the distance between the plates is 3mm3 mm. Calculate the capacitance. If this capacitor is connected to a 100V100 V supply, what is the charge on each plate? (Take Ο΅0=8.854Γ—10βˆ’12F/m\epsilon_0 = 8.854 \times 10^{-12} F/m)

Solution:

  1. Capacitance C=Ο΅0Ad=8.854Γ—10βˆ’12Γ—6Γ—10βˆ’33Γ—10βˆ’3=17.71Γ—10βˆ’12F=17.71pFC = \frac{\epsilon_0 A}{d} = \frac{8.854 \times 10^{-12} \times 6 \times 10^{-3}}{3 \times 10^{-3}} = 17.71 \times 10^{-12} F = 17.71 pF.
  2. Charge Q=CV=17.71Γ—10βˆ’12Γ—100=1.771Γ—10βˆ’9CQ = CV = 17.71 \times 10^{-12} \times 100 = 1.771 \times 10^{-9} C.

Explanation:

We use the standard formula for a parallel plate capacitor in a vacuum and then use the definition of capacitance to find the charge for a given potential.

Problem 2:

Three capacitors of capacitances 2pF2 pF, 3pF3 pF, and 4pF4 pF are connected in parallel. (a) What is the total capacitance of the combination? (b) Determine the charge on each capacitor if the combination is connected to a 100V100 V supply.

Solution:

a) For parallel combination: Cp=C1+C2+C3=2+3+4=9pFC_p = C_1 + C_2 + C_3 = 2 + 3 + 4 = 9 pF. b) In parallel, VV is same for all (100V100 V). Q1=C1V=2Γ—10βˆ’12Γ—100=2Γ—10βˆ’10CQ_1 = C_1 V = 2 \times 10^{-12} \times 100 = 2 \times 10^{-10} C. Q2=C2V=3Γ—10βˆ’12Γ—100=3Γ—10βˆ’10CQ_2 = C_2 V = 3 \times 10^{-12} \times 100 = 3 \times 10^{-10} C. Q3=C3V=4Γ—10βˆ’12Γ—100=4Γ—10βˆ’10CQ_3 = C_3 V = 4 \times 10^{-12} \times 100 = 4 \times 10^{-10} C.

Explanation:

In a parallel circuit, capacitances add up directly, and each component experiences the full source voltage.

Capacitors and Dielectrics - Revision Notes & Key Formulas | ICSE Class 12 Physics