Calculating Potential at Point Above Uniformly Charged Square of Side 2a

[Phymath]

a square of side 2a is uniformly charged with a surface density $$\rho$$
find the potential at a point distance a directly above the midpoint of one of the sides, express your answer as a numerical multiple of $$k_e \rho a$$.

wow ok soo... i got this going for me
$$k \rho \int^a_{-a} \int^a_{-a} \frac{1}{\sqrt{(x-x')^2 + (y-y')^2 + z^2}}dy' dx'$$ where (x,y,z) is the point simple enough..how do i evaulate this...i know it goes all natural log on me...but damn what do i do then!? suggestions anyone?

 

[npgreat]

Find the potential at the point due to a small rod of length 2a and then integrate from a to a+2a

 

[Phymath]

how will that be different? other then a change of limits?

 

[npgreat]

Is the solution
2 ln[4+(2)^1/2]* (KP2a)^2 ?

 

[Phymath]

probley not considering it asks for a numerical multiple of kpa...
no solution is given i just have to figure it out

make it easier..sorta what's this?

$$k \rho \int^a_{-a} \int^a_{-a} \frac{1}{\sqrt{(-x')^2 + (-a-y')^2 + a^2}}dy' dx'$$

 

[Phymath]

how come everytime i do one of these no one answers lol...