CBSE Class 12 · 2004 · ElectrostaticsShort Answer

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State Gauss's theorem in electrostatics. Using this theorem, derive an expression for the electric field intensity due to an infinite plane sheet of charge of charge density

σ C m^{- 2}

Solution

Gauss's theorem — the net flux through any closed surface is

\frac{1}{ε _{0}}

times the charge enclosed:

\oint E \cdot d S = \frac{q _{enc}}{ε _{0}}

Take a cylindrical pillbox of cross-section

A

piercing the sheet symmetrically.

E

is normal to the sheet and equal on both faces; the curved surface carries no flux:

2 E A = \frac{σ A}{ε _{0}}

E = \frac{σ}{2 ε _{0}}

independent of the distance from the sheet.

CBSE Class 12 Board · 2004 · Subjective

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