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Electrostatic Potential and Capacitance - The Parallel Plate Capacitor

Grade 12CBSEPhysics

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

🔑Concepts

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A parallel plate capacitor consists of two large plane parallel conducting plates of area AA separated by a small distance dd.

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The electric field EE between the plates is uniform in the central region and is given by E=σϵ0E = \frac{\sigma}{\epsilon_0}, where σ\sigma is the surface charge density.

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The potential difference VV between the plates is the product of the electric field and the separation distance, represented as V=E⋅dV = E \cdot d.

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Capacitance CC is the ratio of the charge QQ on either plate to the potential difference VV between them, C=QVC = \frac{Q}{V}.

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The capacitance of a parallel plate capacitor in vacuum depends only on geometric factors: C∝AC \propto A and C∝1dC \propto \frac{1}{d}.

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When a dielectric medium with dielectric constant KK completely fills the space between the plates, the capacitance increases to C=KC0C = K C_0, where C0C_0 is the capacitance in vacuum.

📐Formulae

C=ϵ0AdC = \frac{\epsilon_0 A}{d}

E=σϵ0=Qϵ0AE = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}

V=Qdϵ0AV = \frac{Qd}{\epsilon_0 A}

Cmedium=Kϵ0AdC_{medium} = \frac{K \epsilon_0 A}{d}

U=12CV2=Q22CU = \frac{1}{2}CV^2 = \frac{Q^2}{2C}

σ=QA\sigma = \frac{Q}{A}

💡Examples

Problem 1:

A parallel plate capacitor has plates of area A=2×10−2 m2A = 2 \times 10^{-2} \text{ m}^2 and a separation of d=1 mmd = 1 \text{ mm}. Calculate its capacitance in air. (Use ϵ0=8.854×10−12 C2N−1m−2\epsilon_0 = 8.854 \times 10^{-12} \text{ C}^2\text{N}^{-1}\text{m}^{-2})

Solution:

C=ϵ0Ad=8.854×10−12×2×10−21×10−3=1.7708×10−10 FC = \frac{\epsilon_0 A}{d} = \frac{8.854 \times 10^{-12} \times 2 \times 10^{-2}}{1 \times 10^{-3}} = 1.7708 \times 10^{-10} \text{ F}

Explanation:

The capacitance is calculated using the formula C=ϵ0AdC = \frac{\epsilon_0 A}{d}. By substituting the given values for area AA and distance dd (converted to meters), we find the capacitance in Farads.

Problem 2:

A capacitor is initially charged to 750 pC750 \text{ pC}. If an additional charge of 450 pC450 \text{ pC} is added to the plates, what is the final total charge QtotalQ_{total}? Show the calculation using vertical addition.

Solution:

The total charge is the sum of the initial and additional charges: 750+4501200\begin{array}{r} 750 \\ +450 \\ \hline 1200 \end{array} Qtotal=1200 pCQ_{total} = 1200 \text{ pC}

Explanation:

In electrostatics, charges add algebraically. By summing the initial charge and the newly added charge, we obtain the total magnitude of charge on the positive plate.

Problem 3:

If a dielectric slab of dielectric constant K=5K = 5 is inserted between the plates of a 20 \muF20 \text{ \mu F} capacitor, what will be the new capacitance?

Solution:

C′=K×C=5×20 \muF=100 \muFC' = K \times C = 5 \times 20 \text{ \mu F} = 100 \text{ \mu F}

Explanation:

The introduction of a dielectric material between the plates increases the capacitance by a factor equal to the dielectric constant KK of the material.