Gauss’s Law in Electrostatic

1. Concept Overview / Statement of the Law

Gauss’s Law is one of the four fundamental Maxwell’s equations. It connects electric charges to the electric field they produce.

Understanding

Every electric charge produces electric field lines.
If we imagine a closed surface (a balloon, sphere, cube, etc.) around charges, some field lines pass through the surface.

Gauss’s Law tells us:

The total electric flux passing through any closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space.

Mathematically,

[\oint \vec{E} \cdot d\vec{A}] [= \dfrac{Q_{\text{enclosed}}}{\varepsilon_0}]

Here:

• [\oint] means integral over a closed surface
• [\vec{E}] is electric field
• [d\vec{A}] is a small area element on the surface
• [Q_{\text{enclosed}}] is total charge inside the surface
• [\varepsilon_0] is permittivity of free space

Gauss’s Law does not require the surface to be spherical or regular. It works for any shape.

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2. Explanation and Derivation

Electric Flux

Electric flux is:

[\phi = \vec{E} \cdot d\vec{A}] [= EA\cos\theta]

For a closed surface, the total flux is:

[
\oint \vec{E} \cdot d\vec{A}
]

Derivation for a Point Charge at the Center of a Sphere

Let a point charge [+q] be at the center of a sphere of radius [r].

 

Field at every point:

[
E = \dfrac{1}{4\pi\varepsilon_0} \dfrac{q}{r^2}
]

Area of sphere:

[
A = 4\pi r^2
]

Total flux:

[\oint \vec{E} \cdot d\vec{A}] [= E \cdot A] [= \dfrac{1}{4\pi\varepsilon_0}\dfrac{q}{r^2}(4\pi r^2)] [= \dfrac{q}{\varepsilon_0}]

Thus,

[
\boxed{\oint \vec{E} \cdot d\vec{A} = \dfrac{Q}{\varepsilon_0}}
]

This result remains true for any closed surface, even if not spherical.

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3. Dimensions and Units

Dimensions

Electric flux:

[\phi] = [E][A] = \left( \dfrac{MLT^{-3}A^{-1}}{} \right) [L^2] = [ML^3T^{-3}A^{-1}]

Gauss’s Law:

[\oint \vec{E} \cdot d\vec{A} = \dfrac{Q}{\varepsilon_0}]

Unit of electric flux = Newton metre² per coulomb
([N m^2/C])

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4. Key Features

• Valid for any closed surface → spherical, cylindrical, irregular.
• Depends only on charge enclosed, not on charge outside.
• Extremely useful for computing electric fields of symmetrical charge distributions:
  • Spherical symmetry
  • Cylindrical symmetry
  • Planar symmetry
• Reduces complex integrals to simple algebra.
• Works in electrostatics only (charges at rest).

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5. Important Formulas to Remember

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6. Conceptual Questions with Solutions

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7. FAQ / Common Misconceptions

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8. Practice Questions (With Step-By-Step Solutions)

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Q1. A point charge of [+5 μC] is enclosed in a spherical surface. Find the electric flux.

Step 1: Use Gauss’s Law:

[
\phi = \dfrac{Q}{\varepsilon_0}
]

Step 2: Insert values:

[
\phi = \dfrac{5 \times 10^{-6}}{8.85 \times 10^{-12}}
]

[
\phi = 5.65 \times 10^5 , N,m^2/C
]

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Q2. A closed surface encloses +2C and –5C charges. Find net flux.

[Q_{\text{net}}] [= 2 – 5] [= -3C]

[
\phi = \dfrac{-3}{\varepsilon_0}
]

[
\phi = -3.39 \times 10^{11}Nm^2/C
]

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Q3. A cube encloses no charge. What is the net flux?

[\phi = \dfrac{0}{\varepsilon_0}] [= 0]

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Q4. A point charge of [+q] is placed outside a closed box. What is the flux through the box?

[
\phi = 0
]

Because charge outside does not contribute to net flux.

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Q5. A Gaussian surface encloses charge of magnitude [40,nC]. Find flux.

[
\phi = \dfrac{40 \times 10^{-9}}{8.85 \times 10^{-12}}
]

[
\phi = 4.52 \times 10^3  Nm^2/C
]