1

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).

(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).

Open in App

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 d→S3, 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,

∮S→E.→dS=∮S1→E.→dS1+∮S1→E.→dS2+∮S1→E.→dS3

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

= E∫dS1+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.d→x

∫0dW=∫r1r214πϵ0qq0x2dx

W=14πϵ0qq0[−1x]r2r1

=14πϵ0qq0[1r1−1r2]

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 d→S3, 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,

∮S→E.→dS=∮S1→E.→dS1+∮S1→E.→dS2+∮S1→E.→dS3

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

= E∫dS1+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.d→x

∫0dW=∫r1r214πϵ0qq0x2dx

W=14πϵ0qq0[−1x]r2r1

=14πϵ0qq0[1r1−1r2]

25

View More

Join BYJU'S Learning Program

Join BYJU'S Learning Program