1.

consider a uniform electric fieldoversetto E =3xx 10^(3) hatiN//CWhat is the flux of this field througha square of 10cm on a side whose plane is parallel to the yzplane? (b) What is the flux through the same square if if the normal to its plane makes a 60^(@)angle with the x-axis?

Answer»

Solution :Here ` oversetto E =3xx 10^(3)hat I N//C `
As SIDEOF square =10 cm =0.1m, hence surface area `S= ( 0.1 ) ^(2)=0.01 m^(2) `. The plane of surface area being PARALLEL to y= plane, hence `oversetto S = 0.01 hati m^(2) `
` therefore ` Electric flux of the field ` phi_E = oversetto E. oversetto S= (3xx10 ^(3)hati) .(0.01 hati ) =30 N m^(2)C^(-1) `
(b) When normal to the plane of surface area makesan ANGLE of ` 60^(@) `with the x-axis flux is given by
`phi_E=oversetto.oversettoS =E S cos theta =3xx 10^(3)xx 0.01 xx cos 60^(@) `
` "" =15 Nm^(2)C^(-1) `


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