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:
Can a Warlock become Neutral Good?
Today is the Center
What are these boxed doors outside store fronts in New York?
How does strength of boric acid solution increase in presence of salicylic acid?
"You are your self first supporter", a more proper way to say it
Mage Armor with Defense fighting style (for Adventurers League bladeslinger)
Minkowski space
Do VLANs within a subnet need to have their own subnet for router on a stick?
How to write a macro that is braces sensitive?
What does "Puller Prush Person" mean?
Email Account under attack (really) - anything I can do?
How to find program name(s) of an installed package?
What is the offset in a seaplane's hull?
What would happen to a modern skyscraper if it rains micro blackholes?
Arthur Somervell: 1000 Exercises - Meaning of this notation
Is it possible to do 50 km distance without any previous training?
Test if tikzmark exists on same page
How did the USSR manage to innovate in an environment characterized by government censorship and high bureaucracy?
Can I ask the recruiters in my resume to put the reason why I am rejected?
How could an uplifted falcon's brain work?
Why are 150k or 200k jobs considered good when there are 300k+ births a month?
Fencing style for blades that can attack from a distance
What typically incentivizes a professor to change jobs to a lower ranking university?
Problem of parity - Can we draw a closed path made up of 20 line segments...
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:
$begingroup$
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
asked Mar 29 at 15:40
EnlightenedFunkyEnlightenedFunky
84211022
$endgroup$
|
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
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
asked Mar 29 at 15:40
EnlightenedFunkyEnlightenedFunky
84211022
$endgroup$
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
integration physics indefinite-integrals
asked Mar 29 at 15:40
EnlightenedFunkyEnlightenedFunky
84211022
asked Mar 29 at 15:40
EnlightenedFunkyEnlightenedFunky
84211022
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
84211022
$begingroup$
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
$begingroup$
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
$begingroup$
And explanation of the physics and symbols would greatly enable focused answers.
$endgroup$
– avs
Mar 29 at 15:43
$begingroup$
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
$begingroup$
@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
$begingroup$
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
$begingroup$
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
$begingroup$
Do the dimensions agree?
$endgroup$
– marty cohen
Mar 30 at 15:52
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
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 $$E=kint_0^RfracdQ(r)R^2=kfrac 1R^2q_in=kfracQR^2fracR^3a^3=frackQRa^3$$
answered Apr 1 at 18:17
AndreiAndrei
13.5k21230
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3167271%2fis-there-a-calculus-attempt-at-this-question%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $$E=kint_0^RfracdQ(r)R^2=kfrac 1R^2q_in=kfracQR^2fracR^3a^3=frackQRa^3$$
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 $$E=kint_0^RfracdQ(r)R^2=kfrac 1R^2q_in=kfracQR^2fracR^3a^3=frackQRa^3$$
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 $$E=kint_0^RfracdQ(r)R^2=kfrac 1R^2q_in=kfracQR^2fracR^3a^3=frackQRa^3$$
answered Apr 1 at 18:17
AndreiAndrei
13.5k21230
$endgroup$
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 $$E=kint_0^RfracdQ(r)R^2=kfrac 1R^2q_in=kfracQR^2fracR^3a^3=frackQRa^3$$
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
answered Apr 1 at 18:17
AndreiAndrei
13.5k21230
13.5k21230
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3167271%2fis-there-a-calculus-attempt-at-this-question%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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