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Electric Charges and Fields - Gauss’s Theorem and its Applications

Grade 12CBSEPhysics

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

🔑Concepts

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Electric Flux (ΦE\Phi_E): It is defined as the total number of electric field lines passing normally through a given area. Mathematically, ΦE=∫E⃗⋅dA⃗=∫EdAcos⁡θ\Phi_E = \int \vec{E} \cdot d\vec{A} = \int E dA \cos \theta.

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Gauss’s Law: The total electric flux through any closed surface (Gaussian surface) is equal to 1ϵ0\frac{1}{\epsilon_0} times the net charge enclosed by that surface. It is expressed as ∮E⃗⋅dA⃗=qinϵ0\oint \vec{E} \cdot d\vec{A} = \frac{q_{in}}{\epsilon_0}.

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Gaussian Surface: An imaginary closed surface such that the intensity of the electric field at every point on the surface is either constant or zero, making the calculation of flux simpler.

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Field due to an Infinitely Long Straight Wire: For a wire with linear charge density λ\lambda, the electric field at a distance rr is directed radially and its magnitude is inversely proportional to rr.

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Field due to a Uniformly Charged Infinite Plane Sheet: The electric field produced by a sheet with surface charge density σ\sigma is uniform, perpendicular to the plane, and independent of the distance from the sheet.

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Field due to a Uniformly Charged Thin Spherical Shell: For a shell of radius RR and charge qq, the field inside (r<Rr < R) is always zero. Outside (r≥Rr \geq R), the field follows the inverse square law, acting as if the charge is concentrated at the center.

📐Formulae

ΦE=∮E⃗⋅dA⃗=qenclosedϵ0\Phi_E = \oint \vec{E} \cdot d\vec{A} = \frac{q_{enclosed}}{\epsilon_0}

E=λ2πϵ0r(Infinitely long wire)E = \frac{\lambda}{2\pi \epsilon_0 r} \quad \text{(Infinitely long wire)}

E=σ2ϵ0(Infinite plane sheet)E = \frac{\sigma}{2\epsilon_0} \quad \text{(Infinite plane sheet)}

E=14πϵ0qr2(Outside spherical shell, r≥R)E = \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} \quad \text{(Outside spherical shell, } r \geq R\text{)}

E=0(Inside spherical shell, r<R)E = 0 \quad \text{(Inside spherical shell, } r < R\text{)}

σ=qA,λ=qL,ρ=qV(Charge densities)\sigma = \frac{q}{A}, \quad \lambda = \frac{q}{L}, \quad \rho = \frac{q}{V} \quad \text{(Charge densities)}

💡Examples

Problem 1:

A point charge of 17.7 μC17.7 \, \mu C is at the center of a cubical Gaussian surface of side 10 cm10 \, cm. What is the net electric flux through the surface?

Solution:

Given q=17.7×10−6 Cq = 17.7 \times 10^{-6} \, C and ϵ0=8.85×10−12 C2N−1m−2\epsilon_0 = 8.85 \times 10^{-12} \, C^2 N^{-1} m^{-2}. According to Gauss's Law, ΦE=qϵ0\Phi_E = \frac{q}{\epsilon_0}. Substituting the values: ΦE=17.7×10−68.85×10−12=2×106 Nm2C−1\Phi_E = \frac{17.7 \times 10^{-6}}{8.85 \times 10^{-12}} = 2 \times 10^6 \, N m^2 C^{-1}.

Explanation:

The flux through a closed surface depends only on the net charge enclosed and not on the shape or size of the surface (the side of the cube is irrelevant).

Problem 2:

An infinite line charge produces a field of 9×104 N/C9 \times 10^4 \, N/C at a distance of 2 cm2 \, cm. Calculate the linear charge density λ\lambda.

Solution:

We use the formula E=λ2πϵ0rE = \frac{\lambda}{2\pi \epsilon_0 r}. Rearranging for λ\lambda: λ=E⋅2πϵ0r\lambda = E \cdot 2\pi \epsilon_0 r. Given E=9×104 N/CE = 9 \times 10^4 \, N/C and r=0.02 mr = 0.02 \, m. Since 14πϵ0=9×109\frac{1}{4\pi \epsilon_0} = 9 \times 10^9, then 2πϵ0=118×1092\pi \epsilon_0 = \frac{1}{18 \times 10^9}. Thus, λ=(9×104)×0.0218×109=10−7 C/m=0.1 μC/m\lambda = (9 \times 10^4) \times \frac{0.02}{18 \times 10^9} = 10^{-7} \, C/m = 0.1 \, \mu C/m.

Explanation:

The electric field of a line charge decreases linearly with the distance rr from the wire.