What is the Electric Field and Potential of a Non-Uniformly Charged Sphere?

In summary: EdA = Q_encl / E_o = E(4*pi*(R^2)) = (2*pi*p_o*r*(R^2))/E_oSo your integral for (b) is wrong because your dv is wrong. To find the E field inside the sphere draw your Gaussian surface at some arbitrary r inside the sphere. What is Q_{enc}? You know that dQ = \rho dv but what are your limits now?ok so for part b i need the electric field inside and outsideInside..E(4\pir2) = Qencl/Eo
  • #1
joemama69
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Homework Statement


A non-uniformly charged sphere of radius R has a charge density p = p_o(r/R) where p_o is constant and r is the distrance from the center of the spere.

a) find the total charge inside the sphere

b) find the electric field everywhere (inside & outside sphere)

c) find the electric potential difference between teh center of the sphere and a distance r away from the center for both r<R and r>R.

d)Graph the potential as afunction of r

Homework Equations





The Attempt at a Solution



a) Find the total charge in sphere.

p = dQ/dV

Q = integral p dV where dV = 4(pi)R^2dR and it is integrated from 0 to R

Q = integral from 0 to R p_o*(r/R)*4*pi*(R^2)*dR = 2*pi*p_o*r*(R^2)

b) Find electric Field inside & out of sphere

Outisde r>R

Q = 2*pi*p_o*r*(R^2)

[tex]\oint[/tex]EdA = Q_encl / E_o = E(4*pi*(R^2)) = (2*pi*p_o*r*(R^2))/E_o

E = (p_o*r)/2E_o

Inside... r<R

If the rest is correct, can i get a hint on finding the inside as to the outside. this confuses me. it seems the only difference is r<R, but how would that effect the equation.
 
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  • #2
joemama69 said:

Homework Statement


A non-uniformly charged sphere of radius R has a charge density p = p_o(r/R) where p_o is constant and r is the distrance from the center of the spere.

a) find the total charge inside the sphere

b) find the electric field everywhere (inside & outside sphere)

c) find the electric potential difference between teh center of the sphere and a distance r away from the center for both r<R and r>R.

d)Graph the potential as afunction of r

Homework Equations

The Attempt at a Solution



a) Find the total charge in sphere.

p = dQ/dV

Q = integral p dV where dV = 4(pi)R^2dR and it is integrated from 0 to R

Q = integral from 0 to R p_o*(r/R)*4*pi*(R^2)*dR = 2*pi*p_o*r*(R^2)

b) Find electric Field inside & out of sphere

Outisde r>R

Q = 2*pi*p_o*r*(R^2)

[tex]\oint[/tex]EdA = Q_encl / E_o = E(4*pi*(R^2)) = (2*pi*p_o*r*(R^2))/E_o

E = (p_o*r)/2E_o

Inside... r<R

If the rest is correct, can i get a hint on finding the inside as to the outside. this confuses me. it seems the only difference is r<R, but how would that effect the equation.
[tex]dQ = \rho dv = (\frac{\rho_0}{R})r 4\pi r^2 dr[/tex]

So your integral for (a) is wrong because your dv is wrong. To find the E field inside the sphere draw your Gaussian surface at some arbitrary r inside the sphere. What is [tex]Q_{enc}[/tex]? You know that [tex] dQ = \rho dv [/tex] but what are your limits now?
 
  • #3
ok so for part A i integrated and got Q = (4[tex]\pi[/tex]por4)/R and wouldn't that be the same as Qencl because it was the charge in the sphere

Einside = (1/4[tex]\pi[/tex]r2)(Qencl/Eo)

what differes when i find it outside
 
  • #4
joemama69 said:
ok so for part A i integrated and got Q = (4[tex]\pi[/tex]por4)/R and wouldn't that be the same as Qencl because it was the charge in the sphere

Einside = (1/4[tex]\pi[/tex]r2)(Qencl/Eo)

what differes when i find it outside

[tex] Q_{total} = \frac{4\pi\rho_0}{R}\int_{0}^{R}r^3dr =
\frac{4\pi\rho_0}{R}\frac{R^4}{4} = \pi\rho_0R^3 [/tex]

To find charge inside all you do is change your limits from 0 -> R to 0 -> r where r is an arbitrary radius inside the sphere.

p.s learn latex it's not that hard https://www.physicsforums.com/showthread.php?t=8997.
 
  • #5
So if you want the E field outside the sphere, [tex] Q_{enc} = Q_{total} [/tex] since the whole sphere is enclosed with your Gaussian surface.
 
  • #6
ok so for part a i wanted the total charge inside sphere which would be Qenc.. correct?

Part A)

Qencl = [tex]\pi[/tex]por4/R

Then for part B i need the Electric field inside and outside

Inside..

E(4[tex]\pi[/tex]r2) = Qencl/Eo

and then outside i would solve for Qtotal as you did earlier and use that instead of Qencl in the same equation for the inside


why does my pi always end up like an exponent
 
  • #7
joemama69 said:
ok so for part a i wanted the total charge inside sphere which would be Qenc.. correct?

Part A)

Qencl = [tex]\pi[/tex]por4/R

Then for part B i need the Electric field inside and outside

Inside..

E(4[tex]\pi[/tex]r2) = Qencl/Eo

and then outside i would solve for Qtotal as you did earlier and use that instead of Qencl in the same equation for the insidewhy does my pi always end up like an exponent

Use latex and that link I gave you, it is simple.

(a) Your answer is wrong (I showed you the answer in a previous post) but you are right that [tex] Q_{enc}(r) = \frac{\pi\rho_0}{R}r^4 [/tex] --> This equation is the enclosed charge as a function of your radius with r between 0 and R. It is asking for the total charge of the sphere though. So what is the total enclosed charge if the whole sphere is included? It is just that equation but with r = R.

(b) For inside the sphere: Looks good so far and you know what [tex] Q_{enc} [/tex] is :
[tex] Q_{enc}(r) = \frac{\pi\rho_0}{R}r^4 [/tex].

For outside the sphere: You know what [tex] Q_{enc} = Q_{total} [/tex] or in otherwords your answer from (a).
 
  • #8
ok sorr i was confused.. i thought that the charge inside would not include the total sphere

ok now i need to find the potential difference for r<R and r>R
Im confused on the integration limits

r<R do i integrate from 0 to R

r>R i integrate from R to r
 
  • #9
joemama69 said:
ok sorr i was confused.. i thought that the charge inside would not include the total sphere

ok now i need to find the potential difference for r<R and r>R
Im confused on the integration limits

r<R do i integrate from 0 to R

r>R i integrate from R to r

It asks for the potential difference FROM THE CENTER OF THE SPHERE and a distance r away from the center

r<R is that a distance r from the center?

r>R is that from the center of the sphere?
 
  • #10
the exact wording is

Find the electric potential difference between the center of the sphere and a distance r away fr the center for both r<R and r>R. Be sure t indicate where you have chsen yur zero reference potential.

the example in the book finds the potential for r>R and they integrate from ra to rb, setting rb = infinity

there is not really an exaple for r<R so I am lost

is there a rule for doing this, the book is not clear
 
  • #11
joemama69 said:
the exact wording is

Find the electric potential difference between the center of the sphere and a distance r away fr the center for both r<R and r>R. Be sure t indicate where you have chsen yur zero reference potential.

the example in the book finds the potential for r>R and they integrate from ra to rb, setting rb = infinity

there is not really an exaple for r<R so I am lost

is there a rule for doing this, the book is not clear

Well if you are setting your zero potential to be at r = infinity then for r>R you would do the same integration. For r<R you still have to integrate from r = infinity but to a point r inside the sphere. This requires two different integrals because there are two different E fields (one for r>R and one for r<R).
 

FAQ: What is the Electric Field and Potential of a Non-Uniformly Charged Sphere?

What is a non-uniform charged sphere?

A non-uniform charged sphere is a spherical object that has an uneven distribution of electric charge. This means that different parts of the sphere have different amounts of charge, resulting in an overall non-uniform charge distribution.

How does the charge distribution affect the electric field of a non-uniform charged sphere?

The charge distribution of a non-uniform charged sphere affects the electric field in that the electric field strength is not constant at all points on the sphere's surface. The electric field will be stronger at points where the charge is concentrated, and weaker at points where the charge is spread out.

Can a non-uniform charged sphere have a net charge of zero?

Yes, a non-uniform charged sphere can have a net charge of zero if the positive and negative charges are evenly distributed throughout the sphere. However, the charge distribution will still be non-uniform, resulting in variations in the electric field strength.

What is the difference between a non-uniform charged sphere and a uniformly charged sphere?

A uniformly charged sphere has an even distribution of electric charge, meaning that all points on the sphere's surface have the same amount of charge. In contrast, a non-uniform charged sphere has an uneven distribution of charge, resulting in variations in the electric field strength.

How can the electric potential of a non-uniform charged sphere be calculated?

The electric potential of a non-uniform charged sphere can be calculated using the equation V = kQ/r, where k is the Coulomb's constant, Q is the total charge of the sphere, and r is the distance from the center of the sphere. However, for non-uniform charged spheres, this equation must be integrated over the entire surface of the sphere to account for the varying charge distribution.

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