- #1
Phymath
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- 0
a square of side 2a is uniformly charged with a surface density [tex]\rho[/tex]
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 [tex]k_e \rho a[/tex].
wow ok soo... i got this going for me
[tex]
k \rho \int^a_{-a} \int^a_{-a} \frac{1}{\sqrt{(x-x')^2 + (y-y')^2 + z^2}}dy' dx' [/tex] 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?
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 [tex]k_e \rho a[/tex].
wow ok soo... i got this going for me
[tex]
k \rho \int^a_{-a} \int^a_{-a} \frac{1}{\sqrt{(x-x')^2 + (y-y')^2 + z^2}}dy' dx' [/tex] 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?