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Define electric flux. Write its SI units. (b) The electric field components due to a charge inside the cube of side 0.1 marea as shown in E_x =alpha x ,where alpha =500 N//C m, E_Y =0 and E_Z =0. Calculate (i)the flux through the cube , and (ii)the charges inside the cube. |
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Answer» Solution :(i) As `E_y and E_Z ` are zero and E =`E_x = ALPHA x ,` hence electric flux is linked only with TWO faces of the cube lying in y-zplane (i.e., perpendicular to `oversetto (E_x) `) At the position of left face of cube `x= 0.1 m , "hence" E_x = alpha x =500 xx 0.1 = 50 N//C ` and surface area of face `s= (0.1) ^(2)=0.001 m^(2) ` ` therefore "" ` Flux on this face ` phi_1 =E_x s= 50 xx0.01 = 0.5N m^(2) C ^(-1)`(inward ) Again on the opposite face of cube (i.e., right face ) x= 0.2 m and hence ` "" E_x. =alpha x= 500 xx 0.2 = 100 N//C ` Flux on this face `phi_2 =E_x.s =100 xx 0.01 =1 N m^(2) C^(-1) ` (outward ) ` therefore ` Net electric flux through the cube ` phi= 0.5 N m^(2)C^(-1)("inward")+ 1.0N m^(2) C^(-1) ("outward")= +0.5N m^(2)C^(-1)` (outward) (ii)From Gauss.s law ` phi= (q)/( in_0) ` `rArr "" ` Charge inside the cube ` q= phi in_0 =+ 0.5 xx 8.85 xx 10 ^(-12)=+ 4.42 xx 10 ^(-12)C ` |
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