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

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SUMMARY

The discussion focuses on calculating the electric field and potential of a non-uniformly charged sphere with a charge density defined as ρ = ρ₀(r/R). The total charge inside the sphere is derived using the integral Q = ∫ρ dV, yielding Q = (4πρ₀R⁴)/4R. The electric field is determined for both inside (r < R) and outside (r > R) the sphere, with the inside field expressed as E = (ρ₀r)/(2ε₀) and the outside field using the total charge. The potential difference between the center and a point at distance r is also calculated, requiring different integration limits for r < R and r > R.

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  • Understanding of electric fields and potentials in electrostatics
  • Familiarity with Gauss's Law and its application to spherical symmetry
  • Proficiency in calculus, particularly integration techniques
  • Knowledge of charge density functions and their implications in electrostatics
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  • Study the application of Gauss's Law in different geometries, focusing on spherical charge distributions
  • Learn about electric potential calculations for various charge configurations
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This discussion is beneficial for physics students, particularly those studying electromagnetism, as well as educators and professionals involved in electrostatics and related fields.

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 the 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)

\ointEdA = 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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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 the 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)

\ointEdA = 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.
dQ = \rho dv = (\frac{\rho_0}{R})r 4\pi r^2 dr

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 Q_{enc}? You know that dQ = \rho dv but what are your limits now?
 
ok so for part A i integrated and got Q = (4\pipor4)/R and wouldn't that be the same as Qencl because it was the charge in the sphere

Einside = (1/4\pir2)(Qencl/Eo)

what differes when i find it outside
 
joemama69 said:
ok so for part A i integrated and got Q = (4\pipor4)/R and wouldn't that be the same as Qencl because it was the charge in the sphere

Einside = (1/4\pir2)(Qencl/Eo)

what differes when i find it outside

Q_{total} = \frac{4\pi\rho_0}{R}\int_{0}^{R}r^3dr = <br /> \frac{4\pi\rho_0}{R}\frac{R^4}{4} = \pi\rho_0R^3

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.
 
So if you want the E field outside the sphere, Q_{enc} = Q_{total} since the whole sphere is enclosed with your Gaussian surface.
 
ok so for part a i wanted the total charge inside sphere which would be Qenc.. correct?

Part A)

Qencl = \pipor4/R

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

Inside..

E(4\pir2) = 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
 
joemama69 said:
ok so for part a i wanted the total charge inside sphere which would be Qenc.. correct?

Part A)

Qencl = \pipor4/R

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

Inside..

E(4\pir2) = 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 Q_{enc}(r) = \frac{\pi\rho_0}{R}r^4 --> 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 Q_{enc} is :
Q_{enc}(r) = \frac{\pi\rho_0}{R}r^4.

For outside the sphere: You know what Q_{enc} = Q_{total} or in otherwords your answer from (a).
 
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
 
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).
 

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