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Electrostatics - Gauss's Theorem

Grade 12ICSEPhysics

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 crossing a given area. Mathematically, it is the surface integral of the electric field E⃗\vec{E} over a surface SS, given by ΦE=∮SE⃗⋅dA⃗\Phi_E = \oint_S \vec{E} \cdot d\vec{A}.

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Gauss's Theorem Statement: The total electric flux through any closed surface (Gaussian surface) is equal to 1ϵ0\frac{1}{\epsilon_0} times the net charge qenclosedq_{enclosed} trapped inside that surface.

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Gaussian Surface: An imaginary closed surface used to calculate the electric field. For a point charge or spherical distribution, a concentric sphere is used; for a line charge, a co-axial cylinder is used.

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Field due to an Infinitely Long Straight Wire: The electric field EE at a distance rr from a wire with linear charge density λ\lambda is radially outward (if λ>0\lambda > 0).

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Field due to a Uniformly Charged Infinite Plane Sheet: The electric field EE is independent of the distance from the sheet and is given by σ2ϵ0\frac{\sigma}{2\epsilon_0}, where σ\sigma is the surface charge density.

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Field due to a Uniformly Charged Thin Spherical Shell: Outside the shell (r>Rr > R), it behaves like a point charge at the center. Inside the shell (r<Rr < R), the electric field is zero because the enclosed charge is zero.

📐Formulae

ΦE=∮E⃗⋅dA⃗=EAcos⁡θ\Phi_E = \oint \vec{E} \cdot d\vec{A} = E A \cos \theta

Φtotal=qenclosedϵ0\Phi_{total} = \frac{q_{enclosed}}{\epsilon_0}

Ewire=λ2πϵ0rE_{wire} = \frac{\lambda}{2\pi \epsilon_0 r}

Esheet=σ2ϵ0E_{sheet} = \frac{\sigma}{2\epsilon_0}

Eshell=14πϵ0qr2 (for r≥R)E_{shell} = \frac{1}{4\pi \epsilon_0} \frac{q}{r^2} \text{ (for } r \geq R)

Einside=0 (for r<R)E_{inside} = 0 \text{ (for } r < R)

💡Examples

Problem 1:

A point charge q=10μCq = 10 \mu C is placed at the center of a cube of side 10 cm10 \text{ cm}. Calculate the electric flux through each face of the cube.

Solution:

Total flux through the cube Φtotal=qϵ0=10×10−68.854×10−12≈1.13×106 N m2C−1\Phi_{total} = \frac{q}{\epsilon_0} = \frac{10 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 1.13 \times 10^6 \text{ N m}^2\text{C}^{-1}. Since the cube has 66 identical faces and the charge is at the center, the flux through one face is Φface=Φtotal6=1.13×1066≈1.88×105 N m2C−1\Phi_{face} = \frac{\Phi_{total}}{6} = \frac{1.13 \times 10^6}{6} \approx 1.88 \times 10^5 \text{ N m}^2\text{C}^{-1}.

Explanation:

According to Gauss's Law, the total flux depends only on the enclosed charge. Symmetry allows us to divide the total flux equally among the six faces of the cube.

Problem 2:

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

Solution:

Using the formula E=λ2πϵ0rE = \frac{\lambda}{2\pi \epsilon_0 r}, we can rearrange for λ\lambda: λ=E(2πϵ0r)\lambda = E(2\pi \epsilon_0 r). Given E=9×104 N/CE = 9 \times 10^4 \text{ N/C}, r=0.02 mr = 0.02 \text{ m}, and 14πϵ0=9×109 Nm2C−2\frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \text{ Nm}^2\text{C}^{-2}. So, 2πϵ0=12×9×1092\pi \epsilon_0 = \frac{1}{2 \times 9 \times 10^9}. Thus, λ=9×104×0.0218×109=10−7 C/m=0.1μC/m\lambda = 9 \times 10^4 \times \frac{0.02}{18 \times 10^9} = 10^{-7} \text{ C/m} = 0.1 \mu \text{C/m}.

Explanation:

This problem applies the Gauss's Law derivation for a cylindrical symmetry where the electric field is inversely proportional to the distance rr from the line charge.

Gauss's Theorem - Revision Notes & Key Formulas | ICSE Class 12 Physics