Applications of Gauss's Law to Wire, Sheet and Shell
Applications of Gauss's law use symmetry to find electric field due to charge distributions such as an infinitely long line charge, an infinite plane sheet, and a uniformly charged spherical shell.
In one line: Line uses cylinder, sheet uses pillbox, shell uses sphere.
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- Gauss's law is most useful when a Gaussian surface can be chosen so that electric field is either constant and parallel to area vectors, or perpendicular to area vectors and gives zero flux.
- For an infinitely long line charge, a coaxial cylinder is used and E = λ/(2πε0r).
- For an infinite plane sheet, a pillbox surface gives E = σ/(2ε0) on either side.
- For a spherical shell, a concentric sphere shows E = 0 inside the shell and outside field behaves as if total charge were concentrated at the centre.
Example
Formula and rule
Infinite line charge: E = λ/(2πε0r), λ in C m−1, r in m.
Infinite plane sheet: E = σ/(2ε0), σ in C m−2, independent of distance for an ideal infinite sheet.
Spherical shell: E = 0 inside; outside E = (1/4πε0)(Q/r2), as if charge were at centre.
Conditions: ideal symmetry, electrostatic equilibrium where applicable, and suitable Gaussian surface.
See it worked out
Follow the example line by line, then use the same method on new questions.
Worked example
A uniformly charged infinite plane sheet has σ = 3.54 × 10−6 C m−2.
- 1Electric field on either side is E = σ/(2ε0) = (3.54 × 10−6)/(2 × 8.85 × 10−12) = 2.0 × 105 N C−1.
- 2For positive σ, field is normal to the sheet and away from it on both sides.
Method you can reuse
- 1Identify symmetry of charge distribution.
- 2Choose matching Gaussian surface.
- 3Mark direction of E and dA.
- 4Find where E·dA contributes and where it is zero.
- 5Write flux in simplified form.
- 6Put qenclosed in terms of λ, σ, or Q.
- 7Solve for E and state direction and valid region.
Visual tip
Useful diagrams: cylindrical Gaussian surface around line charge, pillbox crossing plane sheet, and concentric spherical Gaussian surfaces inside and outside a charged shell. Labels should show symmetry axis, radius r, area vectors, enclosed charge, and field direction.
Check yourself
Answer the question in your head before opening the answer.
Try it yourself
Why is the electric field inside a uniformly charged spherical shell zero?
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Common wrong answer
Using E = σ/ε0 for a single infinite non-conducting sheet is wrong in this context; the standard result is E = σ/(2ε0) on each side.
Don't mix this up
Exam tip
Write it in the exam
Each point is worth about one mark. Write point 1 for a 1-mark question and keep adding points as the marks go up.
How CBSE asks it
- 1
Enough for 1 mark
Applications of Gauss's law use symmetry to find electric field due to charge distributions such as an infinitely long line charge, an infinite plane sheet, and a uniformly charged spherical shell.
- 2
Add for 2 marks
Example: For a long straight line charge with λ = 4 × 10−6 C m−1 at r = 0.20 m, E = λ/(2πε0r) = (2kλ)/r = (2)(9 × 109)(4 × 10−6)/0.20 = 3.6 × 105 N C−1 radially outward for positive λ.
- 3
Add for 3 marks
Gauss's law is most useful when a Gaussian surface can be chosen so that electric field is either constant and parallel to area vectors, or perpendicular to area vectors and gives zero flux.
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