Derivation of electric potential charged disk from ring

[cookiemnstr510510]

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 R²/y² ) to show that the electric potential now resembles that of a point charge

Homework Equations

V=q/(4πε₀r)

The Attempt at a Solution

a)V=q/(4πε₀r)
dv=dq/(4πε₀)
See ring.jpg attached
V=∫dv=∫dq/(4πε₀)=[1/(4πε₀r)]∫dq=q/4πε₀r=q/4πε₀(R'²+y²)^1/2

b)See disk.jpg attached
V=kq/(R'²+y²)^1/2
dV=kdq/(R'²+y²)^1/2

(dq/Q)=(2πR'dr/πR²)→dq=2QR'dr/R²
The limits on the below integral are from 0→R until stated differently
V=∫dV=∫(k2QR'dr)/[R²(R'²+y²)^1/2]=(kQ/R²)∫2R'dr/(R'²+y²)^1/2
Usub:
U=R'²+y²
dU=2R'dr
dr=dU/2R'
New Limits:
when r=0, U=y²
when r=R, U= R²+y²
after taking integral:
(KQ/R²) [u^1/2]evaulated from y²→R²+y²=(KQ/R²)[(R²+y²)-y]

c) This is the part I am confused on... not sure where to start.
I know if y>>R then my above answer turns into
(KQ/R²)[y²-y] but other than that not sure where to go...

 

[SammyS]

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?
...

Homework Equations

V=q/(4πε₀r)

The Attempt at a Solution

a)V=q/(4πε₀r)
dv=dq/(4πε₀)
See ring.jpg attached

V=∫dv
=∫dq/(4πε₀)
=[1/(4πε₀r)]∫dq
=q/(4πε₀r)
=q/(4πε₀(R'²+y²)^1/2)
.

This result (Part a) looks good. Remember that the charge, Q is the total charge on the disk, whereas, q, will become the portion of the total charge on a thin ring,

Also, you should enclose in parentheses the whole of any denominators in the above expressions. I added those in RED above.

 

[SammyS]

Homework Statement

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.

Homework Equations

V=q/(4πε₀r)

The Attempt at a Solution

V=q/(4πε₀(R'²+y²)^1/2)

b)See disk.jpg attached

V=kq/(R'²+y²)^1/2
dV=kdq/(R'²+y²)^1/2

(dq/Q)=(2πR'dr/πR²)→dq=2QR'dr/R²
The limits on the below integral are from 0→R until stated differently
V=∫dV=∫(k2QR'dr)/[R²(R'²+y²)^1/2]=(kQ/R²)∫2R'dr/(R'²+y²)^1/2

Any r as in dr you have in this part should be R' and dR'

Usub:
U=R'²+y²
dU=2R'dr
dr=dU/2R'

The u-substitution looks fine, but your integration result doesn't look so good.

Also, after finding the antiderivative (indefinite integral) I find it handy to switch back to the original variable(s) and then substitute the limits of integration: Less chance for mistakes.

New Limits:
when r=0, U=y²
when r=R, U= R²+y²

Before integrating, let's see what the integral is !
∫2R'dr/(R'²+y²)^1/2
becomes (without limits for now)
$\displaystyle \int \frac{dU}{\left( U \right)^{1/2}}$
but you might be better off writing it as:

$\displaystyle \int {U ^{-1/2} \ } dU$​

You made an error with the integration.

Added in Edit:
Never mind. the integration looks fine. DUH !

after taking integral:
(KQ/R²) [u^1/2]evaulated from y²→R²+y²=(KQ/R²)[(R²+y²)-y]

Well, that last result should be: $\displaystyle \ V(y)=\frac{2k Q}{R^2}\left( \sqrt{R^2+y^2\,} - y \right)$

(Fixed a missing factor of 2)

 

[cookiemnstr510510]

Well, that last result should be: V(y)=kQR2(√R2+y2−y)

I re did it also and now I am confusing myself:
Im thinking the final answer should be V(y)=(2KQ/R²)[(√R²+y²-y)]
because evaluating the integral:
∫U^-1/2=2U^1/2?

 

[SammyS]

I re did it also and now I am confusing myself:
Im thinking the final answer should be V(y)=(2KQ/R²)[(√R²+y²-y)]
because evaluating the integral:
∫U^-1/2=2U^1/2?

After posting (#3) I did a couple of quick edits.

A couple of mathematical pointers.

Writing ∫U^−1/2 without the dU is a mistake. But, yes, there should be that factor of 2 in the result. (I'll fix my post.)

As for the formula you have for V(y), make sure that the scope of the radical, √ , is clear. Parentheses can help, but it be even clearer to write that with a rational exponent as ( ( R² + y²)^−1/2 − y )

 

[cookiemnstr510510]

After posting (#3) I did a couple of quick edits.

A couple of mathematical pointers.

Writing ∫U^−1/2 without the dU is a mistake. But, yes, there should be that factor of 2 in the result. (I'll fix my post.)

As for the formula you have for V(y), make sure that the scope of the radical, √ , is clear. Parentheses can help, but it be even clearer to write that with a rational exponent as ( ( R² + y²)^−1/2 − y )

appreciate the pointers. Thank you!
Any suggestions on the last part of the problem?

 

[SammyS]

Homework Statement

c)if y>>R, use the binomial expansion (keeping terms up to and including R²/y² ) to show that the electric potential now resembles that of a point charge

Homework Equations

V=q/(4πε₀r)

The Attempt at a Solution

c) This is the part I am confused on... not sure where to start.
I know if y>>R then my above answer turns into
(KQ/R²)[y²-y] but other than that not sure where to go...

Of course we know that last statement is wrong. The potential obtained above is: $\displaystyle \ V(y)=\frac{2k Q}{R^2}\left( \sqrt{R^2+y^2\,} - y \right)$
The first thing to do is to manipulate the portion that depends on $\ y\$ to to have terms with $\ R/y \$.

A reasonably clean way to do this is to define a new (temporary) variable, $\ u\,.\$ Let $\displaystyle \ u = R/y\,,\ \text{ so that } \ R = u\cdot y \,.\$
Notice that if y >> R, then 1 >> u.

You should get $\displaystyle \ \left( \sqrt{R^2+y^2\,} - y \right) = \left( \sqrt{u^2y^2+y^2\,} - y \right) \,.$

Now, factor out $\ y \$ from the expression on the right side of the equation.

"Binomial expansion of what?", you may be asking?

Look at Wikipedia: Newton's generalized binomial theorem .

$\displaystyle \sqrt {x+1} = 1+{\frac {1}{2}}x-{\frac {1}{8}}x^{2}+{\frac {1}{16}}x^{3}-{\frac {5}{128}}x^{4}+{\frac {7}{256}}x^{5}-\cdots$

Plug in $\ u^2 \$ for $\ x\$ and see what you get.