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Question

# An infinitely long straight conductor is uniformly charged with charge density λ per meter. The work done to bring a charge q0 at perpendicular distance b to a perpendicular distance a(a<b) from the conductor is :

A
Zero
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B
q0λ2πε0
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C
q0λ2πε0log0(ba)
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D
q0λ2πε0log0(ab)
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Solution

## The correct option is C q0λ2πε0log0(ba)Charge density =λ/mWork done to being a charge =q0a<bGauss law in electrostatic, it states that total electric flux over the closed surface S is 1ϵ0 times that total charge (q) contained inside S.∮S¯¯¯¯E0d¯¯¯s=qϵ0Electric field due to an infinity long straight wire. Electric field due to an infinity long straight wire.Let us consider an infinity long time charge having linear charge density λ. Assume a cylindrical Gaussian surface of radius r and length l coaxial with the line charge,By symmetry, the electric field E has the same magnitude at each point of the curved surface S1 and is directed radially outward. So, angle at surface between ds1 and E1 is zero and angle of d¯¯¯s2, d¯¯¯s3 with ¯¯¯¯E at S2 and S3 are 900.Total flux through the cylindrical surface,∮¯¯¯¯E.¯¯¯ds=∮S1¯¯¯¯E.d¯¯¯s1+∮S2¯¯¯¯E.d¯¯¯s2+∮S3¯¯¯¯E.d¯¯¯s3∮S1Eds1cos00+∮S2Eds2cos900+∮S3Eds3cos00=E∮d¯¯¯s1=E×2πrlSince λ is the charge per unit length and l is the length of the wire,thus the charge enclosed q=λlAccording to Gaussian law∮¯¯¯¯E¯¯¯¯¯ds=qϵ0or, E×2πrl=λlϵ0E=λ2πϵ0rWork done =−∫baqEdr =−qλa2πϵ0log(ba)

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