IIT JEE , AIEEE Forum - Physics , Electricity

sulekha_hi

Find the flux of the electric field thruogh a spherical surface of radius R due to a charge of 10 to the power -7 at the center and another equal charge at a point 2R away from the center.

b) A charge q is placed at the center of the open cylindrical vessel. the flux of electric field through the surface of the vessel is q/2

how can this be true. the two surfaces are not parallel plate conductor.

neeraj_agarwal_1990

1) the charge kept at a distance of 2r does take part in calculating fulx because flux= Q(enclosed)/ep.

so flux= 10^(-7)/ep.

akumar_ak

for first :

e.f = 10^-7/8.85*10^-12

hope u understand what the denominator means

and it doesnt matter of that another charge

for second :

e.f = E(2*pi*r*l)

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