A disk of radius R (Fig. P25.73) has a nonuniform surface charge density s 5 Cr, where C is a constant and r is measured from th
e center of the disk to a point on the surface of the disk. Find (by direct integration) the electric potential at P.
1 answer:
Answer:
Explanation:
dV = k (dq)/d
dV = k (sigma da)/d
dV = k (5 C r (r dr d(phi)))/d
d = sqrt(r^2 + x^2)
dV = 5 k C (r^2/sqrt(r^2+x^2)) dr d(phi)
==> V = int{5 k C (r^2/sqrt(r^2+x^2)) dr d(phi)} ; from r=0 to r=R and phi=0 to phi=2pi
==> V = 5 k C int{(r^2/sqrt(r^2+x^2)) dr d(phi)} ; from r=0 to r=R and phi=0 to phi=2pi
==> V = 5 k C (2 pi) int{(r^2/sqrt(r^2+x^2)) dr} ; from r=0 to r=R
==> V = 5 k C (2 pi) (1/2) (R sqrt(R^2+x^2) - x^2 ln(sqrt(r^2+x^2) + r) + x^2 Ln(x))
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Explanation:
15N/3cm=5
Convert 3kg to N (29.42N)
Use the 5 from earlier to divide 29.42/5=5.884
Round to 5.9
5.9cm is the answer
(I don’t know if I solved this the correct way but I got it right. 5.9 is the correct answer)