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Given a uniform electric fieldoversetto E =5xx10 ^(3) hati NC^(-1). Find the flux of this field through square of 10 cm on a side whose plane is parallel to the y-Z plane. What would be the flux through the same square if the plane makes a 30^(@)angle with the x-axis? |
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Answer» Solution :Here `oversetto E= 5xx 10^(3) hatiNC^(-1)` and area of square S = 10 `XX 10 cm ^(2)= 100 cm ^(2)= 100 xx 10^(4)m^(2)=0.01 m ^(2) ` Since surface is in Y-Zplane , hence `oversetto S =0.01 hat I m^(2) ` ` therefore "" ` Flux ` phi_in oversetto E. oversetto S= ES =( 5 xx10^(3) ) xx0.01 = 50 N m^(2)C^(-1) ` Whenplane of square makes ` 30 ^(@) ` angle with the x-axis ,then area vector ` oversetto S ` subtends an angle ` theta = 60^(@) ` from x-axis and hence new flux will be `"" phi_in . =ES cos theta = (5xx10^(3) )xx 0.01 xx cos 60^(@) =25 N m^(2) C^(-1) ` |
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