Flux of a point charge through a circle

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

I will try to choose the correct option using the common sense instead of solving it.

As d decreases, the flux should increase. For R>>d, only option (a) and (d) satisfy this condition.

Now, for choosing between (a) and (d),

Let us consider a cylinder of radius R and length 2d placed symmetric about the charge Q.

Now, using the Gauss's law,

$2\int_{S_{circular} } \vec E \cdot d \vec A + \int_{S_{curved}} \vec E \cdot d \vec A = \frac { Q } {\epsilon_0 }$

$\int_{S_{circular} } \vec E \cdot d \vec A = \frac { Q } {2 \epsilon_0 } - \int_{S_{curved}} \vec E \cdot d \vec A$

So, the required flux should be less than $\frac { Q } {2 \epsilon_0 }$.

For R>2d, option (d) does not satisfy the above condition.

So, the correct option is (a).

Is there a way to get the correct option easier than this?

 

[haruspex]

Is there a way to get the correct option easier than this?

You could consider an extreme value of d.

 

[Pushoam]

You could consider an extreme value of d.

The extreme value of d could be either o or infinity. Right?
If I take d as infinity, then I will have to struggle with option a and d.
If I take d near to 0, then using a small cylinder and the abv approach, it gives option a as then I can neglect the flux due to the curved surface..
But, still, I will have to use that cyllinder approach.

Is there any other easier way?

 

[haruspex]

But, still, I will have to use that cyllinder approach.

Isn't it obvious that half the flux goes each way out of the circle?

 

[Pushoam]

Isn't it obvious that half the flux goes each way out of the circle?

For me,
earliier it wasn't .
now it is.
Thanks for it.