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E(r) = Q/4*pi*epsilon-nought(r^2+R^2)

1. the questions asks "why this model/behaviour for r>>R might be "reasonable"

2. what q(r) is needed for Gauss's law to produce the above E(r)

3. what rho(r) in turn is needed to produce this q(r)?

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- #1

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E(r) = Q/4*pi*epsilon-nought(r^2+R^2)

1. the questions asks "why this model/behaviour for r>>R might be "reasonable"

2. what q(r) is needed for Gauss's law to produce the above E(r)

3. what rho(r) in turn is needed to produce this q(r)?

- #2

Hootenanny

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Welcome to Physics Forums.

E(r) = Q/4*pi*epsilon-nought(r^2+R^2)

1. the questions asks "why this model/behaviour for r>>R might be "reasonable"

2. what q(r) is needed for Gauss's law to produce the above E(r)

3. what rho(r) in turn is needed to produce this q(r)?

According to our guidelines (see https://www.physicsforums.com/showthread.php?t=5374"), you are expected to show an attempted solution when asking for help, or at the very least detail your thoughts. This is why we have a homework template, please use it.

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- #3

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1. could it be that because the model assumes that where Gauss's law is used, 2 Gaussian surfaces are separately used and calculated with total charge enclosed over the 2 surfaces still being Q, and hence the (r^2+R^2) instead of just r^2? if that makes any sense. i know that outside the original sphere, for distances r >> R, the relationship between E & r is such that E inversely proportional to r^2. could it be that due to the possibility of that the charge is not uniformly distributed also?

2. As for no 2, using gauss's law, integral of E.dA = Qin/epsilon-nought. not too sure how to get about after though. is it just rearranging every term in terms of Qin?

3. i know its related to no 2 in a way that Qin = rho x Vsphere and via differentiation and integration.

Some thoughts and help here much appreciated.

- #4

Hootenanny

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You're on the right lines. In fact, you hit the nail on the head here:1. could it be that because the model assumes that where Gauss's law is used, 2 Gaussian surfaces are separately used and calculated with total charge enclosed over the 2 surfaces still being Q, and hence the (r^2+R^2) instead of just r^2? if that makes any sense. i know that outside the original sphere, for distances r >> R, the relationship between E & r is such that E inversely proportional to r^2. could it be that due to the possibility of that the charge is not uniformly distributed also?

In the limit r >> R, the expression reduces to the expression for a point charge. In other words, the electric field at a large distance from the nucleus is virtually identical to that of a point charge, as it should be. Do you follow?know that outside the original sphere, for distances r >> R, the relationship between E & r is such that E inversely proportional to r^2.

Indeed, that is it. You will also need to evaluate the integral of course.2. As for no 2, using gauss's law, integral of E.dA = Qin/epsilon-nought. not too sure how to get about after though. is it just rearranging every term in terms of Qin?

The differential form of Gauss' law would be helpful here.3. i know its related to no 2 in a way that Qin = rho x Vsphere and via differentiation and integration.

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So integral of E.da=qin/empslom

if we differentiate both sides we get E= dq/da*(1/epslom).... so we have determined dq/da, which is equal to E*epslom, but how does that allow you to determine the charge density... because wouldnt charge density equal dq/dv.

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Hootenanny

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http://en.wikipedia.org/wiki/Gauss's_law#Differential_form

So integral of E.da=qin/empslom

if we differentiate both sides we get E= dq/da*(1/epslom).... so we have determined dq/da, which is equal to E*epslom, but how does that allow you to determine the charge density... because wouldnt charge density equal dq/dv.

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- #8

Hootenanny

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No, that it not Gauss' law, is it? The charge density is equation to the permittivity of free space (epsilon zero) multiplied by theOh yh. So the charge density is simply equal to epslom multiplied by the electric field??

P.S. Are you Boon28?

- #9

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Can u explain it in a simple way?

P.S I'm not warrior_1 :/

- #10

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integral of E.dA = Q/epsilon-zero

solving the integral and rearranging, I got q(r) = Qr^2/(r^2+R^2)..

for no. 3: charge density i was told i can get by finding the derivative of q(r) with respect to r.

and so i got rho(r) = dq/dr = 2QrR^2/(r^2+R^2)^2

Do those seem reasonable?

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