- #1

cookiemnstr510510

- 162

- 14

## Homework Statement

a)recall that electric potential is a scalar quantity. For a circular ring of radius, R', carrying charge, Q, what is the electric potential at a height,y, above the center of the ring?

b)Use your above answer to determine the electric potential at a height,y, above the center of a uniformly charged disk of radius, R.

c)if y>>R, use the binomial expansion (keeping terms up to and including R

^{2}/y

^{2}) to show that the electric potential now resembles that of a point charge

## Homework Equations

V=q/(4πε

_{0}r)

## The Attempt at a Solution

a)V=q/(4πε

_{0}r)

dv=dq/(4πε

_{0})

See ring.jpg attached

V=∫dv=∫dq/(4πε

_{0})=[1/(4πε

_{0}r)]∫dq=q/4πε

_{0}r=q/4πε

_{0}(R'

^{2}+y

^{2})

^{1/2}

b)See disk.jpg attached

V=kq/(R'

^{2}+y

^{2})

^{1/2}

dV=kdq/(R'

^{2}+y

^{2})

^{1/2}

(dq/Q)=(2πR'dr/πR

^{2})→dq=2QR'dr/R

^{2}

The limits on the below integral are from 0→R until stated differently

V=∫dV=∫(k2QR'dr)/[R

^{2}(R'

^{2}+y

^{2})

^{1/2}]=(kQ/R

^{2})∫2R'dr/(R'

^{2}+y

^{2})

^{1/2}

Usub:

U=R'

^{2}+y

^{2}

dU=2R'dr

dr=dU/2R'

New Limits:

when r=0, U=y

^{2}

when r=R, U= R

^{2}+y

^{2}

after taking integral:

(KQ/R

^{2}) [u

^{1/2}]evaulated from y

^{2}→R

^{2}+y

^{2}=(KQ/R

^{2})[(R

^{2}+y

^{2})-y]

c) This is the part Im confused on... not sure where to start.

I know if y>>R then my above answer turns into

(KQ/R

^{2})[y

^{2}-y] but other than that not sure where to go...