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.

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))

i wanna say it’s between b and d, leaning more towards b.

answer: d. a cme increases the amount of energy in the solar wind

explanation:

coronal mass ejections, also called cmes, are large clouds of plasma and magnetic field that erupt from the sun. only when the cloud is aimed at earth will the cme hit earth and therefore cause impacts. high-speed solar wind streams come from areas on the sun known as coronal holes.