Electric field and potential of a sphere

In summary, the conversation discussed the calculation of electric field and potential for a sphere with a constant charge density. The total charge of the sphere was determined using a triple integral, and the electric field was found to be radial outside the sphere and zero inside. Some mistakes were made in the initial calculations, but they were corrected with the help of others in the conversation.
  • #1
fluidistic
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Homework Statement



A sphere of radius [tex]a[/tex] has a distribution of charge [tex]\rho (r)=Cr[/tex] where [tex]C[/tex] is a constant.
1)Calculate the electric field in all points of space.
2)Calculate the electric potential outside the sphere.

Homework Equations



To figure out.

The Attempt at a Solution


I got all wrong, I'd like to know where are my mistakes.
1) The electric field is radial. To calculate it outside the sphere I first calculate the total charge [tex]Q[/tex] of the sphere : [tex]Q= \int _0^a \rho (r) dr = \int _0 ^a Crdr= \frac{Ca^2}{2}[/tex].
Now I imagine a sphere (Gaussian surface) of radius [tex]r>a[/tex] and I have that [tex]\Phi=\frac{Q_{\text{enclosed}}}{\varepsilon _0} = \oint \vec E d \vec A=EA=E4 \pi r^2[/tex] but we already saw that [tex]Q_{\text{enclosed}}= \frac{Ca^2}{2}[/tex], thus [tex]E4 \pi r^2 = \frac{Ca^2}{2 \varepsilon _0} \Rightarrow E= \frac{Ca^2}{8 \pi \varepsilon _0 r^2}[/tex], [tex]r>a[/tex]. (wrong result)

Now I calculate the electric field inside the sphere : [tex]\Phi=\oint \vec E d \vec A = 4 \pi r^2 E = \frac{Cr^2}{2 \varepsilon _0} \Rightarrow E= \frac{C}{8 \pi \varepsilon _0}[/tex] for [tex]r<a[/tex]. (wrong result, as you can see I got that the electric field inside the sphere does not depend on [tex]r[/tex]... impossible!)

2)[tex]E=- \nabla \varphi[/tex].
Thus [tex]\varphi (r)=- \int _a ^r E(r)dr = - \int _a ^r \frac{Ca^2}{8 \pi \varepsilon _0 r^2}dr=\frac{Ca^2}{8 \pi \varepsilon _0} \left [ \frac{1}{r}- \frac{1}{a} \right ][/tex], with [tex]r>a[/tex] where [tex]r[/tex] is the distance between the center of the sphere and the considered point outside the sphere.
Which is also wrong.
 
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  • #2
the total charge on the sphere is the surface of the sphere times the charge density so:

Qtot = rho * 4 pi r^2

so in the limit of space the electric field is just :

E = 1/ (4*pi *eta) * Qtot / r^2

however for points close to the sphere this is more complicated and you need some integration.

the electric field inside the sphere however is not a function of r because it is 0 everywhere inside the sphere.

from gauss's law we know that the electric flux = Qinclosed / eta = (integral) E dA

when we draw imaginary spheres inside the sphere, the Q inclosed is zero because all the charge is on the sphere, while we are in it. So because the area of the imaginary sphere is non zero, the electric field HAS to be zero to conform to gauss's law.

Qinclosed = 0 -> electric flux = 0 / eta = 0 = E * A (A is non-zero) -> E = 0
 
  • #3
latrocinia said:
the total charge on the sphere is the surface of the sphere times the charge density so:

Qtot = rho * 4 pi r^2

so in the limit of space the electric field is just :

E = 1/ (4*pi *eta) * Qtot / r^2

however for points close to the sphere this is more complicated and you need some integration.

the electric field inside the sphere however is not a function of r because it is 0 everywhere inside the sphere.

from gauss's law we know that the electric flux = Qinclosed / eta = (integral) E dA

when we draw imaginary spheres inside the sphere, the Q inclosed is zero because all the charge is on the sphere, while we are in it. So because the area of the imaginary sphere is non zero, the electric field HAS to be zero to conform to gauss's law.

Qinclosed = 0 -> electric flux = 0 / eta = 0 = E * A (A is non-zero) -> E = 0

I don't think so. I haven't been clear. The sphere is not empty as the exercise points out that [tex]\rho (r)=Cr[/tex]. Hence [tex]Q[/tex] is not worth [tex]4 \pi a^2 \rho[/tex]. Furthermore the electric field inside the sphere is not null.
 
  • #4
First off, something seems off about your calculation of the total charge. A sphere is a three dimensional object, and to calculate the total charge, you will need the volume. Thus, the integral must be three dimensional, whereas yours is one dimensional.

Why is the electric field at all points in space outside of the sphere not simply the same electric field formula for a point charge?

Take a Gaussian surface of radius r > a. The flux inside is equal to E*SA = Q_inside / eps_0. Q_inside is simply the total charge of the sphere. Change eps_0 into the Coulomb's constant equivalent, and you can see the behavior I mentioned above.

I would help with the inside the sphere, but I'm working on my own homework at the moment. xD Sorry. Here's a little bit of help: E should be directly proportional to r in your final formula.
 
  • #5
TwoTruths said:
First off, something seems off about your calculation of the total charge. A sphere is a three dimensional object, and to calculate the total charge, you will need the volume. Thus, the integral must be three dimensional, whereas yours is one dimensional.

Why is the electric field at all points in space outside of the sphere not simply the same electric field formula for a point charge?

Take a Gaussian surface of radius r > a. The flux inside is equal to E*SA = Q_inside / eps_0. Q_inside is simply the total charge of the sphere. Change eps_0 into the Coulomb's constant equivalent, and you can see the behavior I mentioned above.

I would help with the inside the sphere, but I'm working on my own homework at the moment. xD Sorry. Here's a little bit of help: E should be directly proportional to r in your final formula.
Oh thanks... you're right. Hopefully my only error was with calculating the total charge of the sphere. I hope my reasoning is right otherwise.
 
  • #6
I used calculus III to calculate the total charge of the sphere, which gave me [tex]Q=C \int _0^{2\pi} \int _0^{\pi} \int _0^a r^3 \sin (\phi) d r d\phi d\theta =\pi a^4 C[/tex].
I guess I'm almost done for the exercise.
Thanks a lot.
Thread solved.
 
  • #7
Good work on figuring out the triple integral btw. =)
 
  • #8
fluidistic said:
I used calculus III to calculate the total charge of the sphere, which gave me [tex]Q=C \int _0^{2\pi} \int _0^{\pi} \int _0^a r^3 \sin (\phi) d r d\phi d\theta =\pi a^4 C[/tex].
Good! But take advantage of radial symmetry to get the answer quicker (with calculus I) by viewing the sphere as composed of concentric shells:

[tex]Q = \int _0^a 4\pi r^2 \rho(r) dr = 4\pi C \int _0^a r^3 dr = \pi a^4 C[/tex]
 
  • #9
so if you knew the electric field, how can you derive the charge of the enclosed sphere
 
  • #10
i was thinking using gauss law for next flux = q(enclosed)/eta
 

1. What is an electric field and potential of a sphere?

An electric field is a physical field that surrounds a charged object and exerts a force on other charged objects within its range. The electric potential of a sphere is the amount of electric potential energy that a unit charge would have if placed at a certain point within the electric field of the sphere.

2. How is the electric field and potential of a sphere calculated?

The electric field and potential of a sphere can be calculated using the equations E = kQ/r^2 and V = kQ/r, where k is the Coulomb constant, Q is the charge of the sphere, and r is the distance from the center of the sphere.

3. What factors affect the electric field and potential of a sphere?

The electric field and potential of a sphere are affected by the charge of the sphere, the distance from the center of the sphere, and the surrounding medium (such as air or water).

4. How does the electric field and potential change as you move away from the sphere?

The electric field and potential decrease as you move further away from the sphere, following the inverse square law. This means that the field strength and potential decrease exponentially as the distance increases.

5. What are some real-life applications of understanding the electric field and potential of a sphere?

Understanding the electric field and potential of a sphere is important in many practical applications, such as designing electrical systems, calculating the forces on charged particles in a plasma, and predicting the behavior of charged particles in planetary and solar systems.

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