IIT JEE , AIEEE Forum - Physics , Electricity

imnotserious_2008

1. a point charge Q is located on the axis of a disc of radius R at a distance a from the plane of the disc.if one fourth(1/4 th) of the flux from the charge passes through the disc, then find the relation between a and R.

2. A charge Q is uniformly distributed over a rod length L. CONSIDER A HYPOTHETICAL CUBE OF EDGE L WITH THE centre of the cube AT ONE END of the rod. find the minimum possible flux of the electric field through the entire surface of the cube.

ans: 1.a=R/3^1/2

2. Q/(2episilon not)

jatinroxx

The flux passing thru a surface that subtends an angle 2@ at the point is given by

1/2 *(1- cos@)q/

₀)

This must be equal to q/4

₀

We get cos @ as pi/3

Now that the angle is known, the relation can be identified.

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

krishna.gopal

Here i am giving general method to solve this problem. On the disc make an ring of radius r and thickness dr.
Electric field at any point onthis ring = kQ/(r^2+a^2) and it makes an agle of cos inverse (a/sqrt(a^2+r^2)) with the normal to disc.So flux through this ring

d

= (2*pi*r*dr)*(K*Q*a)/(a²+r²)^3/2

Now integrate it from 0 to R, you will get flux through disc and hence the relation.

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