(adsbygoogle = window.adsbygoogle || []).push({}); 1. A sphere has a charge distrubution such that the charge density goes as P(r)=Ar^(1/2). Where A=0.200 C/ M^(7/2). If the sphere has a radius of 2.00m, determine the electric potential at a point 3.00m from the center of the sphere.

I understand that P=Q/V (Im using P as Rho) but because the charge is not constant i must intergrate it. So P(r)= dQ/dV, then dQ=P(r)dV

V of sphere= ((4pi)r^3)/3) So dV= (4pi)r^2 and therefore dQ=P(r)(4pi)r^2

Now i beleive that A and R are constants so I removed them from the intergral. After intergrating both sides of the equation i get Q=(Ar^(1/2)) (((4pi)r^2)/3)

I plugged in all the numbers and came out with 0.945 for Q. The answer for Q should be 0.812.

Can anyone tell me where I went wrong?

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# Homework Help: Charge distrubution in a sphere

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