Example – 1 If a point charge Q is located at the centre of a cube then find (i) flux through the total
surface, of the cube (ii) flux through one surface.
Solution : (i) According to Gauss’s law for closed surface Æ = Q/e0 , (ii) Since cube is a symmetrical figure thus by symmetry the flux through each surface is Æ ‘ = Q/6e0
Example – 2 A point charge +q is located L/2 above the centre of a square having side L. Find the flux through this square.
Solution : The charge q can be supposed to be situated at the centre of a cube having side L with outward flux Æ. In this case the square is one of its face having flux Æ/6
\ flux though the square = (q/6e0)
Example – 3 What is the value of electric flux in SI unit in Y-Z plane of area 2m2, if intensity of electric field is ®E = (5i^ + 2i^) N/C.
Solution: Æ = ®.®dA
= (5i^ + 2j^).2i^
= 10 V-m
Example – 4 2mC charge is in some Gaussion surface, given outward flux Æ, what additional charge is needed if we want that 6Æ flux enters into the Gaussian surface.
Solution : According to question
Q = - 14mC
Example – 5 A cylinder of length L and radius b has its axis coincident with the x axis. The electric
field in this region ®E = 200i^ . Find the flux through (a) the left end of cylinder (b) the right end of cylinder (c) the cylinder curved surface, (d) the closed surface area of the cylinder.
Solution : from fig. then
(a) Æa = ®E.®A = EAcosq
= 200 x pb2 x cos p
=- 200 pb2
(b) Æb = EA cos 0°
= 200 pb2
(c) Æc = EA cos 90° = 0
(d) Æ = Æa + Æb + Æc
= - 200 pb2 + 200 pb2 + 0 = 0
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