Q24 of 26 Page 1

(a) Define electric flux. Is it a scalar or a vector quantity ?

A point charge q is at a distance of d/2 directly above the centre of a square of side d, as shown in the figure. Use Gauss’ law to obtain the expression for the electric flux through the square.



(b) If the point charge is now moved to a distance ‘d’ from the centre of the square and the side of the square is doubled, explain how the electric flux will be affected.


OR


(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).

a) The electric flux through a given area held inside an electric field is the measure of the total number of electric lines of force passing normally through that area. It is a scalar quantity.


Electric flux ∅E = ∫E.dS


Where E = electric field


dS = area element


We can imagine the square as face of a cube with edge d and with charge q placed at its centre, as shown in figure below.



According to Gauss’s theorem, the total flux of the electric field E through a closed surface S is equal to 1/ϵo times the charge q enclosed by the surface S.



Symmetry of six faces of a cube about its centre ensures that the flux ∅s through each square face is same when the charge is placed at the centre.


∴ total flux,




Hence the flux through square of side d with charge q at a distance d/2 directly above the head is q/6ϵo.


b) If the charge is moved to a distance d and the side of the square is doubled the cube will be constructed to have a side 2d but the total charge enclosed in it will remain the same. Hence the total flux through the cube and therefore the flux through the square will remain the same as before.


OR


a) Electric field (E) due to a straight uniformly charged infinite line of charge density λ C/m.



Consider a thin long straight wire having a uniform linear charge density λ C/m. By symmetry, the field E of the line is directed radially outwards and its magnitude is same at all points equidistant from the line charge. To determine the field at a distance r from the line charge we choose a cylindrical gaussian surface of radius r, length l and with its axis along the line charge. As shown in figure, it has a curved surface S1 and a flat circular surface S2 and S3. From figure,


dS1||E


dS2and dS3 are perpendicular to E.


so, only the curved surface contributes towards the total flux.




As cos90° = 0 and cos0° = 1, total flux will be


∅E = E∫dS1 = E× area of the curved surface


∅E = E× 2πrl


Charge enclosed by gaussian surface , q = λl.


Using gauss theorem, we get




b) Graph the variation of E with perpendicular distance r from the line of charge is shown below:



c) Work done in moving the charge q. Through a small displacement dr


dW = F.dr


Now electric F is given as


F = qE


Where E = electric field


∴ dW = qEdr cos0° = qE dr


Substituting the value of E from part (a) we get



Work done in moving the given charge from

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OR


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