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An electric field is uniform and in the positive x direction for positive x and uniform with the same magnitude but in the negative x direction for negative x. It is given that E = 200 hati N//Cand E = - 200 hatiN/C for x lt0. A right circular cylinder of length 20 cm and radius 5 cm has its centre at the origin and its axis along the *-axis so that one face is at jc = + 10 cm and the other is at x = - 10 cm as shown in figure, (a) What is the net outward flux through each flat face ? (b) What is the flux through the side of the cylinder ? (c) What is the net outward flux through the cylinder ? (d) What is the net charge inside the cylinder ? |
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Answer» Solution :`A_(1) = A_(2) = pir^(2)` `=(3.14) (0.05)^(2)` `=7.85 xx 10^(-3) m^(2)` (a) ELECTRIC flux linked with face-1 is, `phi_(1) =A_(1)E_(1) cos theta_(1)` `=(7.85 xx 10^(-3)) (200) cos0^(@)` `therefore phi_(1) = 1.57 (Nm^(2))/C` Electric flux linked with face-2 is, `ph_(2) = A_(2)E_(2) cos theta_(2)` `=(7.85 xx 10^(-3)) (200) cos 0^(@)` `therefore phi_(2) = 1.57 (Nm^(2))/C` (b) Here curved surface of given cylinder is parallel to electric field and so no electric flux PASSES through this curved surface. Hence, `phi_(3)=0` (c) NET electric flux passing through given cylinder is, `phi= phi_(1) + phi_(2) + phi_(3)` `=1.57 + 1.57 + 0` `therefore phi = 3.14 (Nm^(2))/C` (d) If q is the net charge enclosed by above cylinder then according to Gauss theorem, `phi = q/(epsilon_(0))` `therefore q = phi epsilon_(0)` `=(3.14)(8.85 xx 10^(-12))` `therefore q = 2.779 xx 10^(-11) C` |
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