Electrostatics potential calculation for a uniformly charged square

[AHSAN MUJTABA]

Homework Statement

Consider charge distributed on the surface of a square of sides with length
2s that are parallel to the x and y axes. The square is in the xy plane. The surface
charge density is assumed to take the form of y^n for n = 0, 1, 2... . Elucidate the
behavior of the electric potential for r >>s and show that you get the correct behavior
by computing the exact potential on points on the x axis.

Relevant Equations

the potential due to surface charge distribution

I took a surface element dA at the surface of square at point x',y' now I took a point on x-axis and calculated the flux. But I got a very complicated integral though it should be simple and I can't interpret it

 

[haruspex]

It is unclear where the centre of the square should be, and r is undefined.
Based on the fact that the side length is given as 2s, I suspect that the square is supposed to be centred on the origin.
Maybe you are being asked for the potential at (r,0) for r>>s.

Wrt posting images, they are supposed to be for diagrams and textbook extracts. Please take the trouble to type in equations, preferably in latex.
And try to ensure there are no shadows across images. Your diagram is quite hard to read.

 

[AHSAN MUJTABA]

I took my square in xy plane as asked and I took 'r' to be the separation vector from my 'dq' i.e charge element in dA to the field point on x axis.

 

[haruspex]

I took my square in xy plane as asked

Sure, but that does not specify where in the xy plane. It does not say first quadrant, and it does not specify the origin to be at one corner.

I took 'r' to be the separation vector from my 'dq'

Why? Nothing in the problem statement says what r represents.

Are there more details in the original question, lost in translation perhaps?

 

[AHSAN MUJTABA]

no actually there are no more details

 

[haruspex]

no actually there are no more details

One more clue... it says the sides are parallel to the x and y axes. If it intended that two sides lay on those axes I feel it would have said so, so that's another reason for assuming the square is centred on the origin.

 

[AHSAN MUJTABA]

ok so if square is on the origin then normally what we do is we take a point above the origin let's say on z axis, and we calculate potential there. Can we do the same for point on the x axis? if yes then the potential should be zero along the whole x-axis because of consideration of -ve x-axis as well?

 

[haruspex]

ok so if square is on the origin then normally what we do is we take a point above the origin let's say on z axis, and we calculate potential there. Can we do the same for point on the x axis? if yes then the potential should be zero along the whole x-axis because of consideration of -ve x-axis as well?

Even with the square centred on the origin, the integral is extremely nasty.

The potential will indeed be zero on the x-axis when n is odd, but not when n is even. But that's not to do with the negative x axis, it's to do with the antisymmetry about the x axis.
Maybe just try the odd n case first.

 

[AHSAN MUJTABA]

can you please elaborate antisymmetry in this case please?

 

[AHSAN MUJTABA]

and also I wanted to ask that the integral expression I wrote is right or wrong? If just change the limits from -s to s.

 

[haruspex]

can you please elaborate antisymmetry in this case please?

If n is odd, how does the potential at (s,0) due to the charge on the plate at (x,y) relate to the potential at (s,0) due to the charge on the plate at (x,-y)?