Electric field due to charged sphere.

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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 im 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 im 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 im using the right method)
 
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Answers and Replies

  • #2
gabbagabbahey
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then im 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
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yeah meant to put that :) ive 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 dont get anywhere.
 
  • #4
gabbagabbahey
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yeah meant to put that :) ive 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 dont 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
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where did that epslion nought come from in the denominater?
 
  • #6
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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 im a multiple of 2 to big in the denominater.... cant see hwere that happend...
 
  • #7
gabbagabbahey
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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 im a multiple of 2 to big in the denominater.... cant see hwere that happend...
Are you sure? Remember the final answer has an 8 in the denominator:wink:
 
  • #8
gabbagabbahey
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where did that epslion nought come from in the denominater?
Sorry, it was a typo.
 
  • #9
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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
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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 ive gone wrong...
 
  • #11
gabbagabbahey
Homework Helper
Gold Member
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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 ive 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
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Oooops i had root 9 written on my piece of paper, must of typed it out as just 9 :P
yeah im pretty sure i was right as its just a simple derivitive, thanks again mate :)
 
  • #13
gabbagabbahey
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Welcome :smile:
 

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