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Is there a calculus attempt at this question:

Given a continuous charge distribution of a sphere calculate the electric field away:Why $ intrhodV = frac1c intj^0dV = frac1c intj^idS_i $?Finding a scalar field whose gradient is a given conservative vector fieldFormally evaluating integral to calculate electric or gravitational field.Continuity of the divergence of a static electric fieldElectric field lines between surfaces of hollow spherePotential of a charged sphere.Proportionality, integrationHow to do this problem without using infinitesimal?Gauss's law in infinite spaceGiven a continuous charge distribution of a sphere calculate the electric field away:

0

1

$begingroup$

This question is based on the electric field.
This is the question which I would like to solve using this integral:
This is the equation for Electric Field.
$$E=int fracdqr^s$$
I was wondering if this approach was valid?

This is volume charge density and an exression can developed that states:

$$dq=rho dV$$
$$dq = rho 4pi r^2 dr$$
$$rho = fracq_inV_in=fracQ_totalV_total$$

Which leads me to say then that:

$$fracq_infrac43pi a^3cdotfrac43 pi r^3=Q$$
Then when I take the derivative of both sides I get:

$$frac3q_inr^2a^3dr=dQ$$

$$kint frac3q_inr^2r^2a^3dr$$
$$E=3kfracq_ina^3r$$

integration physics indefinite-integrals

share|cite|improve this question

edited Mar 30 at 15:42

EnlightenedFunky

asked Mar 29 at 15:40

EnlightenedFunkyEnlightenedFunky

84211022

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And explanation of the physics and symbols would greatly enable focused answers.
  $endgroup$
  – avs
  Mar 29 at 15:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @avs Did you read the last question?
  $endgroup$
  – EnlightenedFunky
  Mar 29 at 15:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, don't have time to follow links or look into sources. (Didn't even notice your link.)
  $endgroup$
  – avs
  Mar 29 at 16:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Questions should be self-contained. It is alright to reference another question to give more background, but you should reproduce enough here to allow people to understand your question without having to follow the link. However, you've added several symbols to your calculation that are not defined here, nor are they defined in the other question. So we have to guess what it is you mean. Sorry, not interested in guessing.
  $endgroup$
  – Paul Sinclair
  Mar 30 at 1:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do the dimensions agree?
  $endgroup$
  – marty cohen
  Mar 30 at 15:52
  
  

  
  
  
  

 | 
show 1 more comment

0

1

$begingroup$

This question is based on the electric field.
This is the question which I would like to solve using this integral:
This is the equation for Electric Field.
$$E=int fracdqr^s$$
I was wondering if this approach was valid?

This is volume charge density and an exression can developed that states:

$$dq=rho dV$$
$$dq = rho 4pi r^2 dr$$
$$rho = fracq_inV_in=fracQ_totalV_total$$

Which leads me to say then that:

$$fracq_infrac43pi a^3cdotfrac43 pi r^3=Q$$
Then when I take the derivative of both sides I get:

$$frac3q_inr^2a^3dr=dQ$$

$$kint frac3q_inr^2r^2a^3dr$$
$$E=3kfracq_ina^3r$$

integration physics indefinite-integrals

share|cite|improve this question

edited Mar 30 at 15:42

EnlightenedFunky

asked Mar 29 at 15:40

EnlightenedFunkyEnlightenedFunky

84211022

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And explanation of the physics and symbols would greatly enable focused answers.
  $endgroup$
  – avs
  Mar 29 at 15:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @avs Did you read the last question?
  $endgroup$
  – EnlightenedFunky
  Mar 29 at 15:53
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Sorry, don't have time to follow links or look into sources. (Didn't even notice your link.)
  $endgroup$
  – avs
  Mar 29 at 16:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Questions should be self-contained. It is alright to reference another question to give more background, but you should reproduce enough here to allow people to understand your question without having to follow the link. However, you've added several symbols to your calculation that are not defined here, nor are they defined in the other question. So we have to guess what it is you mean. Sorry, not interested in guessing.
  $endgroup$
  – Paul Sinclair
  Mar 30 at 1:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do the dimensions agree?
  $endgroup$
  – marty cohen
  Mar 30 at 15:52
  
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

This question is based on the electric field.
This is the question which I would like to solve using this integral:
This is the equation for Electric Field.
$$E=int fracdqr^s$$
I was wondering if this approach was valid?

This is volume charge density and an exression can developed that states:

$$dq=rho dV$$
$$dq = rho 4pi r^2 dr$$
$$rho = fracq_inV_in=fracQ_totalV_total$$

Which leads me to say then that:

$$fracq_infrac43pi a^3cdotfrac43 pi r^3=Q$$
Then when I take the derivative of both sides I get:

$$frac3q_inr^2a^3dr=dQ$$

$$kint frac3q_inr^2r^2a^3dr$$
$$E=3kfracq_ina^3r$$

integration physics indefinite-integrals

share|cite|improve this question

edited Mar 30 at 15:42

EnlightenedFunky

asked Mar 29 at 15:40

EnlightenedFunkyEnlightenedFunky

84211022

$endgroup$

share|cite|improve this question

edited Mar 30 at 15:42

EnlightenedFunky

asked Mar 29 at 15:40

EnlightenedFunkyEnlightenedFunky

84211022

share|cite|improve this question

edited Mar 30 at 15:42

EnlightenedFunky

asked Mar 29 at 15:40

EnlightenedFunkyEnlightenedFunky

84211022

asked Mar 29 at 15:40

EnlightenedFunkyEnlightenedFunky

84211022

asked Mar 29 at 15:40

EnlightenedFunkyEnlightenedFunky

84211022

1 Answer
1

active

oldest

votes

1

$begingroup$

The formula next to last is not correct. Let's say that the original sphere has radius $a$, and we want to calculate the electric field at some distance $R$ from the center. A spherical shell that you use for integration has a radius $r$, with $0le rle R$. The electric field due to this shell is $fracdQR^2$, and not $fracdQr^2$ as you have used in your calculations. Then, you get $$E=kint_0^RfracdQ(r)R^2=kfrac 1R^2q_in=kfracQR^2fracR^3a^3=frackQRa^3$$

share|cite|improve this answer

answered Apr 1 at 18:17

AndreiAndrei

13.5k21230

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add a comment | 

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1

$begingroup$

The formula next to last is not correct. Let's say that the original sphere has radius $a$, and we want to calculate the electric field at some distance $R$ from the center. A spherical shell that you use for integration has a radius $r$, with $0le rle R$. The electric field due to this shell is $fracdQR^2$, and not $fracdQr^2$ as you have used in your calculations. Then, you get $$E=kint_0^RfracdQ(r)R^2=kfrac 1R^2q_in=kfracQR^2fracR^3a^3=frackQRa^3$$

share|cite|improve this answer

answered Apr 1 at 18:17

AndreiAndrei

13.5k21230

$endgroup$

add a comment | 

1

$begingroup$

The formula next to last is not correct. Let's say that the original sphere has radius $a$, and we want to calculate the electric field at some distance $R$ from the center. A spherical shell that you use for integration has a radius $r$, with $0le rle R$. The electric field due to this shell is $fracdQR^2$, and not $fracdQr^2$ as you have used in your calculations. Then, you get $$E=kint_0^RfracdQ(r)R^2=kfrac 1R^2q_in=kfracQR^2fracR^3a^3=frackQRa^3$$

share|cite|improve this answer

answered Apr 1 at 18:17

AndreiAndrei

13.5k21230

$endgroup$

add a comment | 

$begingroup$

The formula next to last is not correct. Let's say that the original sphere has radius $a$, and we want to calculate the electric field at some distance $R$ from the center. A spherical shell that you use for integration has a radius $r$, with $0le rle R$. The electric field due to this shell is $fracdQR^2$, and not $fracdQr^2$ as you have used in your calculations. Then, you get $$E=kint_0^RfracdQ(r)R^2=kfrac 1R^2q_in=kfracQR^2fracR^3a^3=frackQRa^3$$

share|cite|improve this answer

answered Apr 1 at 18:17

AndreiAndrei

13.5k21230

$endgroup$

share|cite|improve this answer

answered Apr 1 at 18:17

AndreiAndrei

13.5k21230

answered Apr 1 at 18:17

AndreiAndrei

13.5k21230

answered Apr 1 at 18:17

AndreiAndrei

13.5k21230