IIT JEE , AIEEE Forum - Physics , Electricity

Avinash_Bhat

1. The total electric flux over any Gaussian surface enclosing charge Q is : Q /

₀

SO : Q / 4

₀ = 1/4th the flux =

E.dA = E

R² ( Area of the disk is

R² )

SO : E = Q / 4

₀R² ----------------- (1)

BUT : E at the axis for the disk is given by : ( 2Q / 4

₀R² ) * ( 1 - a /

(a² + R²) ) ------------------ (2)

FROM (1) and (2) : a = R /

3

2. Given the rod is having a total charge of Q. So half the rod has Q/2 charge.

Only half the rod is inside the cube. SO : As per Gauss's law , the flux over the cubical surface is 1/

₀times the total charge enclosed by the cube which is Q/2. SO : Flux over the entire surface is Q/2

₀

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