Calculating Electrostatic Potential for a Uniformly Charged Sphere

[Fazza3_uae]

How To Find Potential ??

Homework Statement

Consider the uniformly charged sphere with
radius R = 4.65461 m,
Q = 6.66731 μC is the
total charge
inside the
sphere.

a) Find the total flux passing through the
Gaussian surface (a spherical shell) with ra-
dius 2.08762 m.

b) Find the electric field at radius r =
2.08762 m.

c) Find the value of the electrostatic potential
at the same radius r = 2.08762 m.

Homework Equations

E=kq/r^2
v=kq/r

The Attempt at a Solution

I have answered Parts A & B , I got for part A : Total Flux = (Q/Epsilon)*(r/R)^3
& for part B : E = kQr/R^3 , But part C have stopped me so any hint or clue will be gratefull and thanks in advance.

 

[tiny-tim]

Welcome to PF!

Hi Fazza3_uae! Welcome to PF!

(try using the X² tag just above the Reply box )

I have answered Parts A & B , I got for part A : Total Flux = (Q/Epsilon)*(r/R)^3
& for part B : E = kQr/R^3 , But part C have stopped me so any hint or clue will be gratefull and thanks in advance.

[STRIKE](But isn't the total flux independent of radius?)
[/STRIKE]
What definition do you know for electric potential?

 

[Fazza3_uae]

Hi Fazza3_uae! Welcome to PF!

Hi Tiny-Tim! Thx 4 Welcoming ^^

(But isn't the total flux independent of radius?)

Please check my steps of solving this part & tell me if i made a mistake :


$$\phi=\int\vec{E}.d\vec{A}$$

= Int(EdACos[theta]) = Int(EdA) = EA


E : Electric Field inside a uniform sphere of charge = KQr/R³

A : Area of the spherical Shell of radius r = 4*Pi*r²

Inserting these Values we get :

$$\phi=$$ (Q/epsilon)(r/R)³

What definition do you know for electric potential?

As far as i knew from the lectures the electric potential is the potential energy divided by the charge of a point in space.I also do know that Difference in potential is the same as Flux.

If i understood it wrongly please correct it for me & let me know & it will be appreciated. ^^

 

[tiny-tim]

Hi Fazza3_uae!

(have a pi: π and an epsilon: ε )

(But isn't the total flux independent of radius?)

oops! ignore that … I misread the numbers.

I also do know that Difference in potential is the same as Flux.

No …

i] when we say "potential", we always mean the difference in potential anyway (and the same for "potential energy")

ii] Flux is the integral of a vector across a surface.

(In particular, there is no such thing as flux "at a point".)

So electric flux is the integral of the electric field (E) across a surface.

(Read http://en.wikipedia.org/wiki/Flux#Flux_definition_and_theorems_2" )

So the way you answer a) (the flux) is simply by measuring the charge inside the surface, and dividing by ε₀ …

then you answer b) (the field) by using your formula flux = EA.

As far as i knew from the lectures the electric potential is the potential energy divided by the charge of a point in space.

It would be more accurate to say that electric potential = potential energy per charge.

But for c) you need an equation for the potential (or for the potential energy) … what equation do you know?

 

[Fazza3_uae]

Hi again sir ^.*

Thx a lot for the information and corection that U've posted & I appreciate your help to me Sir.

Part C ) MmmMmmm... I actually know an equation for potential which is V=kq/r but I tried it and got wrong answer. I've tried for r once and for R once & still got wrong answer.

I doubt if there is another formula of potential that is related to the Formula I got in part B. I mean it could be like this : V=kqr/R

I know that My doubt is Maybe 1% correct but still a try. I can't check if this new equation is right or not because I have tried to put an answer twice in the website & still one chance to insert my last answer otherwise i'll get Zero in this part.

Thank you again sir. ^^

( Please Excuse Me If My English Is So Bad, I will Try To Improve It )

 

[tiny-tim]

Part C ) MmmMmmm... I actually know an equation for potential which is V=kq/r but I tried it and got wrong answer. I've tried for r once and for R once & still got wrong answer.

I doubt if there is another formula of potential that is related to the Formula I got in part B. I mean it could be like this : V=kqr/R

I know that My doubt is Maybe 1% correct but still a try. I can't check if this new equation is right or not because I have tried to put an answer twice in the website & still one chance to insert my last answer otherwise i'll get Zero in this part.

Thank you again sir. ^^

( Please Excuse Me If My English Is So Bad, I will Try To Improve It )

oh i see … you used v for the potential.

Yes, if E = kq/r², then E = -gradU, so U (or v) = -kq/r.

 

[Fazza3_uae]

You said if E=kq/r², then E=-gradU , What do u mean by (gradU) & why it is minus ?

and here we have E=kqr/R³, so what V must equal now??

 

[tiny-tim]

You said if E=kq/r², then E=-gradU , What do u mean by (gradU) & why it is minus ?

If you haven't done grad, then just read it as (d/dr) … so E = -dV/dr.

And it's minus because that's how the potential is defined.

(For example, the field outside the sphere is K/r², and the potential is -K/r, so that the potential is 0 at r = ∞ )

and here we have E=kqr/R³, so what V must equal now??

oh, I forgot that E=kqr/R³ … ok, then dV/dr = -kqr/R³, so V = … ?

 

[Fazza3_uae]

Yup , Finally I got right answer Thx a lot Sir.

& thanks also for the clarification about grad ^^ Appreciate it.

May God Bless You Sir.

 

[tiny-tim]

thanks …

but no need to call me Sir …

i'm only a little goldfish! ​

 

[Fazza3_uae]

Alright Tiny Miny Little GoldFish U got it ^^