Question

calculate the flux through the flat disc held perpendicular to the end of a semi infinite wire of charge per unit length lamda. radius of disc is R

Answer 1

I can't see a simple way of doing this, using Gauss Law and symmetry.
Here, the electric field at all points at a distance 'r' from the centre of the disc on the disc must be equal by symmetry.
Now, by integration you can find the value of that electric field. Just take small elements along the wire and find their electric field at the required point. Now take only one component of this field, the one perpendicular to the disc, because the other component will give zero in the dot product taken while finding the flux. Do the required integration over the entire wire.
Now, divide the disc into small rings, radius varying from zero to the radius of the disc. Now, calculate flux for each of these = area of the thin ring element * the component of the electric field at a point at a distance equal to the radius of the element from the centre of the disc. Now, integrate it over the disc. This should give you your answer !!!

Answer 2

Is it \(\Phi_E=\Large \frac{\lambda R}{2\epsilon _0}\) ?

Answer 3

@Vincent-Lyon.Fr - it is \[2(pi)k \lambda R\]

Answer 4

Coulomb's constant K is can be written as:
\( K=\Large \frac{1}{4\pi \epsilon _0} \)
so both expressions are identical.

Answer 5

E.S = \[k \lambda/R * pi R ^{2}\]
=\[k \lambda pi R\]

Answer 6

@Vincent-Lyon.Fr - you there?

Answer 7

yes, but I do not understand what you wrote.

Answer 8

electric field * Area of the surface = flux

Answer 9

But you cannot multiply because E is not uniform. You have to integrate across the whole disc from r = 0 to r = R.

Answer 10

okay in that case i will get it...

Answer 11

Do it, and if it does not succeed, I'll write a derivation and post it.