CBSE Class 12 · 2005 · All India · ElectrostaticsShort Answer

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State Gauss's theorem. Apply this theorem to obtain the expression for the electric field intensity at a point due to an infinitely long, thin, uniformly charged straight wire.

Solution

Gauss's theorem:

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

Take a coaxial cylindrical surface of radius

r

, length

ℓ

around the wire.

E

is radial and constant over the curved surface; the end faces carry no flux:

E (2 π r ℓ) = \frac{λ ℓ}{ε _{0}}

E = \frac{λ}{2 π ε _{0} r}

directed radially outward for positive

λ

CBSE Class 12 Board · 2005 · All India · Subjective

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