Calculating Electric Flux Through a Square: Is Gauss' Law the Only Method?

[stunner5000pt]

This is SUPPOSED to be easy but i seemingly find find it hard...

A poin charge of +Q is places a distance d/2 above the centre of a square surface of side d. Find the electric flux through the square.

so i know that

E dA = EA (because the flux through the square is all at 90 degree angles) = kQd^2 / (d/2)^2 = Q / pi epsilon0

But the answer is Q / 6epsilon0

have i got the concept wrong here?

please do help!

 

[Doc Al]

so i know that

E dA = EA (because the flux through the square is all at 90 degree angles) = kQd^2 / (d/2)^2 = Q / pi epsilon0

The electric field at the surface is not simply kQ/(d/2)^2: the distance to the surface is not just d/2! Furthermore, the electric field is not perpendicular to that surface! (The field from a point charge radiates out from the center.) To calculate the flux directly, you need to find the component of the field perpendicular to the surface and integrate.

But don't do that. Instead, take advantage of symmetry. Hint: Imagine other sides were added forming a cube around the point charge. (It is easy.)

 

[stunner5000pt]

I think i figured something out, if flux is Qenc / permittivity

then the charge +Q in a cube of side d is simply Qenc / permit

But sinc this is a square, the flux is one sixth (since a cbe has six sides) of what a cube is so it is Q / 6permit

am i right??

 

[Tide]

Why would you think the electric field is everywhere normal to the surface?

 

[stunner5000pt]

Why would you think the electric field is everywhere normal to the surface?

i thought wrong, read my second post, i believe it is more relevant

 

[Doc Al]

I think i figured something out, if flux is Qenc / permittivity

then the charge +Q in a cube of side d is simply Qenc / permit

But sinc this is a square, the flux is one sixth (since a cbe has six sides) of what a cube is so it is Q / 6permit

am i right??

Yes, you are right.

 

[Muhammad Usma]

can't we derive it by any other method?