Electric field due to charged sphere.

In summary, the conversation discussed the problem of calculating the electric field inside and outside a charged sphere using Gauss' theorem. It was noted that the charge density on the sphere could be integrated in spherical polar coordinates to find the total charge, and then used to calculate the electric field inside and outside the sphere. The conversation also mentioned a mistake made in the calculation of the maximum electric field, which was corrected to be E_max = a * sqrt(5)/9. Finally, it was noted that the answer provided in the solutions manual was incorrect, as r should equal sqrt(5)/3 instead of sqrt(3)/9.
  • #1
Spoony
77
0
1. The problem statement, all variables and given known data &

Homework Equations

&

The Attempt at a Solution



the charge density on a sphere readius a is
[tex] \rho = \rho_{0} ( 1 - \frac{r^{2}}{a^{2}} ) [/tex]intergrating this in spherical polar co-ords gives me:

[tex] Q = \frac{8 \pi \rho_{0} a^{3}}{15} [/tex]then I am asked to calculate E inside and outside the sphere in terms of Q, using gauss' theorem

[tex] \int dA . E = Q / \epsilon_{0} [/tex]setting this up for outside with intergration limits a and 0 in all cases (so charge on sphere above used for q) and total area being [tex]4 \pi r^{2} [/tex]
gives me:

[tex] E = \frac{Q}{4 \pi \epsilon_{0} r^{2}} [/tex]
all well and good so far :)

then inside when r < a: this is giving me trouble as the awnser is:
[tex] E = \frac{Q}{8 \pi \epsilon_{0} a^{3}} \left[ {5r} - \frac{3r^{3}}{a^{2}} \right] [/tex]but i canrt seem to get this no matter what i do, my notes on a similar question incourage me to use the results of intergration of the Q charge distribution to get:
[tex] Q = 4 \pi \rho_{0} \left[ \frac{r^{3}}{3} - \frac{r^5}{5a^{2}}\right] [/tex] (1)as q depends on r, therefore e would depend on r. but i then set:
[tex] \int dA . E = \frac{Q}{\epsilon_{0}} [/tex]
with Q = (1) and total area at point r [tex] 4 \pi r^{2} [/tex]
i then get all stuck as I am meant to express it in terms of Q and i end up with nothing like the awnser (note the question asks me to calculate E inside sphere before E outside sphere so i think I am using the right method)
 
Last edited:
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  • #2
Spoony said:
then I am asked to calculate E inside and outside the sphere in terms of Q, using gauss' theorem

[tex] \int dA . E = Q / \epsilon_{0} [/tex]

Careful; you should have

[tex]\int_{\mathcal{GS}} \mathbf{E} \cdot \mathbf{da}=\frac{q_{enc}}{\epsilon_0}[/tex]

Where [itex]q_{enc}[/itex] is the amount of charge enclosed by the gaussian surface [itex]\mathcal{GS}[/itex]. [itex]Q[/itex] is the total charge, and inside the sphere, the charge enclosed will be less than [itex]Q[/itex].

Try calculating [itex]q_{enc}[/itex] in terms of the radius of your gaussian surface and then use your expression for [itex]Q[/itex] to arrive at the desired result.
 
  • #3
yeah meant to put that :) I've tried putting in [tex]q_{enc}[/tex] for surface smaller than a which is:
[tex]
Q = 4 \pi \rho_{0} \left[ \frac{r^{3}}{3} - \frac{r^5}{5a^{2}}\right]
[/tex]

where r is distance from centre of sphere and a a constant (here total radius of sphere), but i don't get anywhere.
 
  • #4
Spoony said:
yeah meant to put that :) I've tried putting in [tex]q_{enc}[/tex] for surface smaller than a which is:
[tex]
Q = 4 \pi \rho_{0} \left[ \frac{r^{3}}{3} - \frac{r^5}{5a^{2}}\right]
[/tex]

where r is distance from centre of sphere and a a constant (here total radius of sphere), but i don't get anywhere.

Okay, so
[tex]
q_{enc} = 4 \pi \rho_{0} \left[ \frac{r^{3}}{3} - \frac{r^5}{5a^{2}}\right]
[/tex]

try factoring out a [tex]\frac{2a^3}{15\epsilon_0}[/tex] from the quantity in brackets. That gives you something of the form:

[tex]q_{enc}=\frac{8\pi\rho_o a^3}{15\epsilon_0}[...]=Q[...][/tex]
 
  • #5
where did that epslion nought come from in the denominater?
 
  • #6
Ok so i get
[tex] Q(r) = Q \left[ \frac{15 r^{3}}{6 a^{3}} - \frac{15 r^5}{10 a^{5}}\right]
[/tex]
im guessing i then have to divide this by area:
[tex]
4 \pi r^{2}
[/tex]
and finally divide by epsilon nought.
look good?

it appears I am a multiple of 2 to big in the denominater... can't see hwere that happend...
 
  • #7
Spoony said:
Ok so i get
[tex] Q(r) = Q \left[ \frac{15 r^{3}}{6 a^{3}} - \frac{15 r^5}{10 a^{5}}\right]
[/tex]
im guessing i then have to divide this by area:
[tex]
4 \pi r^{2}
[/tex]
and finally divide by epsilon nought.
look good?
Yup, what does that give you for E?

it appears I am a multiple of 2 to big in the denominater... can't see hwere that happend...

Are you sure? Remember the final answer has an 8 in the denominator:wink:
 
  • #8
Spoony said:
where did that epslion nought come from in the denominater?

Sorry, it was a typo.
 
  • #9
SUCCESS! thanks very much mate :) Once again youve helped me see what was staring me right in the face, which is what this type of physics really is.
you try over and over in different ways and its always something that was right there at the start and always something simple..ish :P
 
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  • #10
im asked to calculate where the field is maximal, which is when the derivitive of the field with respect to distance is at 0.
where r>a the derivitive gets very close to 0 ut never reaches it.
With r<a the derivitive is:
[tex] \frac{dE}{dr} = \frac{Q}{8 \pi \epsilon_{0} a^{3}} \left[5 - \frac{9 r^{2}}{a^{2}\right]} [/tex]


Setting the part in brackets equal to zero:

[tex] 5 - \frac{9 r^{2}}{a^{2}} = 0 [/tex]


i then obtain that:

[tex] E_{max} = a \frac{\sqrt{5}}{9} [/tex]

But the awnser given in the solutions is:


[tex] E_{max} = a \frac{\sqrt{3}}{9} [/tex]

cant see where I've gone wrong...
 
  • #11
Spoony said:
im asked to calculate where the field is maximal, which is when the derivitive of the field with respect to distance is at 0.
where r>a the derivitive gets very close to 0 ut never reaches it.
With r<a the derivitive is:
[tex] \frac{dE}{dr} = \frac{Q}{8 \pi \epsilon_{0} a^{3}} \left[5 - \frac{9 r^{2}}{a^{2}\right]} [/tex]


Setting the part in brackets equal to zero:

[tex] 5 - \frac{9 r^{2}}{a^{2}} = 0 [/tex]


i then obtain that:

[tex] E_{max} = a \frac{\sqrt{5}}{9} [/tex]

But the awnser given in the solutions is:


[tex] E_{max} = a \frac{\sqrt{3}}{9} [/tex]

cant see where I've gone wrong...

I think both you and the solutions manual are wrong this time:

[tex] 5 - \frac{9 r^{2}}{a^{2}} = 0 \implies r^2=\frac{5}{9}a^2 \implies r=\frac{\sqrt{5}}{3}a[/tex]
 
  • #12
Oooops i had root 9 written on my piece of paper, must of typed it out as just 9 :P
yeah I am pretty sure i was right as its just a simple derivitive, thanks again mate :)
 
  • #13
Welcome :smile:
 

What is an electric field?

An electric field is a physical quantity that describes the influence of an electric charge on other charges in its vicinity. It is a vector field, meaning it has both magnitude and direction.

How is the electric field due to a charged sphere calculated?

The electric field due to a point charge is given by the equation E = kQ/r^2, where E is the electric field, k is a constant, Q is the charge of the sphere, and r is the distance from the center of the sphere. For a charged sphere, this equation can be modified to account for the distribution of charge over the surface of the sphere.

What is the direction of the electric field due to a charged sphere?

The electric field due to a charged sphere is radial, meaning it points away from or towards the center of the sphere. The direction depends on whether the sphere is positively or negatively charged.

How does the electric field due to a charged sphere vary with distance?

The electric field due to a charged sphere follows an inverse square law, meaning it decreases with the square of the distance from the center of the sphere. This means that the electric field is strongest near the surface of the sphere and decreases as you move further away.

Can the electric field due to a charged sphere be shielded?

Yes, the electric field due to a charged sphere can be shielded by placing a conducting material between the sphere and the point of interest. This material will redistribute the charges and neutralize the electric field.

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