Question

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 the center of the disk to a point on the surface of the disk. Find (by direct integration) the electric potential at P.

Answer 1

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))