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Question

(a) Use Gauss' law to derive the expression for the electric field (E) due to a straight uniformly charged infinite line of charge density λ C/m.
(b) Draw a graph to show the variation of E with perpendicular distance r from the line of charge.
(c) Find the work done in bringing a charge q from perpendicular distance r1 to r2(r2>r1).

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Solution

(a) Electric field E due to a straight uniformly charged infinite line of charge density λ : Consider a cylindrical Gaussian surface of radius r and length l coaxial with line charge. The cylindrical Gaussian surface may be divided into three parts : (1) curved surface S1 (ii) flat surface S2 and (iii) flat surface S3.
By symmetry, the electric field has the same magnitude E at each point of curved surface S1 and is directed radially outward. We consider small elements of surfaces S1,S2 and S3. The surface element vector d S1, is directed along the direction of electric field (i.e., angle between E and d S1, is 0); the elements d S2, and dS3, are directed perpendicular to field vector E (i.e., angle between d S2 and E , and d S3, and E is 90).

Electric flux through the cylindrical surface,
SE.dS=S1E.dS1+S1E.dS2+S1E.dS3

= S1 Ed S1 cos 00+S1 Ed S2 cos 900+S1 Ed S3 cos 900

= EdS1+0+0

= E × area of curved surface = E×2πrl
Charge enclosed, q=λl

By Gauss' theorem, ϕE=qϵ0

E. 2πrl=λlϵ0

or E=λ2πϵ0r


(b) Graph showing variation of E with perpendicular distance from line of charge: The electric field is inversely proportional to distance 'r' from line of charge. Thus,

(c) Work done in bringing a charge q from perpendicular distance r1 to r2 (r2>r1), in field of a charge

dW = F.dx

0dW=r1r214πϵ0qq0x2dx

W=14πϵ0qq0[1x]r2r1

=14πϵ0qq0[1r11r2]

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