Potential by Assembling Charges The 2019 Stack Overflow Developer Survey Results Are InPotential difference between Earth's surface and 2 meters abovePotential of a uniformly charged hollow sphereElectric potential inside a conductorElectric field and electric scalar potential of two perpendicular wiresboundary condition of electrical fieldElectric Potential due to Sphere when cavity is at arbitrary positionSystem of point charges, Potential related questionIs this process to compute the electrostatic potential energy a valid one?Do charges move to the outer surface of a conductor to minimize the potential energy?Can Potential Energy be found by Energy Density?
How to make payment on the internet without leaving a money trail?
How long do I have to send my income tax payment to the IRS?
What do hard-Brexiteers want with respect to the Irish border?
Deadlock Graph and Interpretation, solution to avoid
What spell level should this homebrew After-Image spell be?
If the Wish spell is used to duplicate the effect of Simulacrum, are existing duplicates destroyed?
How to manage monthly salary
Are USB sockets on wall outlets live all the time, even when the switch is off?
Are there any other methods to apply to solving simultaneous equations?
Time travel alters history but people keep saying nothing's changed
Inflated grade on resume at previous job, might former employer tell new employer?
Is three citations per paragraph excessive for undergraduate research paper?
How to change the limits of integration
Which Sci-Fi work first showed weapon of galactic-scale mass destruction?
Adding labels to a table: columns and rows
How is radar separation assured between primary and secondary targets?
The difference between dialogue marks
Spanish for "widget"
Access elements in std::string where positon of string is greater than its size
What does Linus Torvalds mean when he says that Git "never ever" tracks a file?
What is the use of option -o in the useradd command?
Where does the "burst of radiance" from Holy Weapon originate?
Does it makes sense to buy a new cycle to learn riding?
How are circuits which use complex ICs normally simulated?
Potential by Assembling Charges
The 2019 Stack Overflow Developer Survey Results Are InPotential difference between Earth's surface and 2 meters abovePotential of a uniformly charged hollow sphereElectric potential inside a conductorElectric field and electric scalar potential of two perpendicular wiresboundary condition of electrical fieldElectric Potential due to Sphere when cavity is at arbitrary positionSystem of point charges, Potential related questionIs this process to compute the electrostatic potential energy a valid one?Do charges move to the outer surface of a conductor to minimize the potential energy?Can Potential Energy be found by Energy Density?
$begingroup$
For finding electric potential energy of a uniformly charged sphere, we can assemble the sphere by brining charges from infinity to that point. So to make a uniformly charged sphere of radius $R$ and total charge $Q$, at some instant, charge will be assembled up to a certain radius $x$.
In order to find potential of this sphere at the surface, why is my approach giving different answers?
Approach 1:
$$rho = frac3Q4 pi R^3$$
$$q = frac43 pi x^3 rho = Q fracx^3R^3$$
Potential at the surface would be $$V = fracq4 pi epsilon_0 x = fracQ x^24 pi epsilon_0 R^3$$
Approach 2:
$$rho = frac3Q4 pi R^3$$
$$q = frac43 pi x^3 rho = Q fracx^3R^3$$
$$E = fracQ x4 pi epsilon_0 R^3$$ (From Gauss' Law)
Potential at the surface would be $$V = -intvecE cdot vecdx = -fracQ4 pi epsilon_0 R^3 int_0^xxdx = -fracQ x^28 pi epsilon_0 R^3$$
Why is the answer different in both the cases?
electrostatics potential
asked 1 hour ago
Kushal T.Kushal T.
416
$endgroup$
add a comment |
$begingroup$
For finding electric potential energy of a uniformly charged sphere, we can assemble the sphere by brining charges from infinity to that point. So to make a uniformly charged sphere of radius $R$ and total charge $Q$, at some instant, charge will be assembled up to a certain radius $x$.
In order to find potential of this sphere at the surface, why is my approach giving different answers?
Approach 1:
$$rho = frac3Q4 pi R^3$$
$$q = frac43 pi x^3 rho = Q fracx^3R^3$$
Potential at the surface would be $$V = fracq4 pi epsilon_0 x = fracQ x^24 pi epsilon_0 R^3$$
Approach 2:
$$rho = frac3Q4 pi R^3$$
$$q = frac43 pi x^3 rho = Q fracx^3R^3$$
$$E = fracQ x4 pi epsilon_0 R^3$$ (From Gauss' Law)
Potential at the surface would be $$V = -intvecE cdot vecdx = -fracQ4 pi epsilon_0 R^3 int_0^xxdx = -fracQ x^28 pi epsilon_0 R^3$$
Why is the answer different in both the cases?
electrostatics potential
asked 1 hour ago
Kushal T.Kushal T.
416
$endgroup$
add a comment |
$begingroup$
For finding electric potential energy of a uniformly charged sphere, we can assemble the sphere by brining charges from infinity to that point. So to make a uniformly charged sphere of radius $R$ and total charge $Q$, at some instant, charge will be assembled up to a certain radius $x$.
In order to find potential of this sphere at the surface, why is my approach giving different answers?
Approach 1:
$$rho = frac3Q4 pi R^3$$
$$q = frac43 pi x^3 rho = Q fracx^3R^3$$
Potential at the surface would be $$V = fracq4 pi epsilon_0 x = fracQ x^24 pi epsilon_0 R^3$$
Approach 2:
$$rho = frac3Q4 pi R^3$$
$$q = frac43 pi x^3 rho = Q fracx^3R^3$$
$$E = fracQ x4 pi epsilon_0 R^3$$ (From Gauss' Law)
Potential at the surface would be $$V = -intvecE cdot vecdx = -fracQ4 pi epsilon_0 R^3 int_0^xxdx = -fracQ x^28 pi epsilon_0 R^3$$
Why is the answer different in both the cases?
electrostatics potential
asked 1 hour ago
Kushal T.Kushal T.
416
$endgroup$
For finding electric potential energy of a uniformly charged sphere, we can assemble the sphere by brining charges from infinity to that point. So to make a uniformly charged sphere of radius $R$ and total charge $Q$, at some instant, charge will be assembled up to a certain radius $x$.
In order to find potential of this sphere at the surface, why is my approach giving different answers?
Approach 1:
$$rho = frac3Q4 pi R^3$$
$$q = frac43 pi x^3 rho = Q fracx^3R^3$$
Potential at the surface would be $$V = fracq4 pi epsilon_0 x = fracQ x^24 pi epsilon_0 R^3$$
Approach 2:
$$rho = frac3Q4 pi R^3$$
$$q = frac43 pi x^3 rho = Q fracx^3R^3$$
$$E = fracQ x4 pi epsilon_0 R^3$$ (From Gauss' Law)
Potential at the surface would be $$V = -intvecE cdot vecdx = -fracQ4 pi epsilon_0 R^3 int_0^xxdx = -fracQ x^28 pi epsilon_0 R^3$$
Why is the answer different in both the cases?
electrostatics potential
electrostatics potential
asked 1 hour ago
Kushal T.Kushal T.
416
asked 1 hour ago
Kushal T.Kushal T.
416
edited 26 mins ago
Kushal T.
asked 1 hour ago
Kushal T.Kushal T.
416
asked 1 hour ago
Kushal T.Kushal T.
416
asked 1 hour ago
Kushal T.Kushal T.
416
416
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Approach 2 is wrong. You didn't take into account the corresponding limits for potential. Potential at centre of sphere is not zero!! The expression is V(x)-V(0) instead of V(x).... Find potential at surface by integrating for electric field outside sphere from X to infinity V(infinity)=0. So Then if you wish you can find V(x) by integrating from x=x to any general x=y(
answered 1 hour ago
TojrahTojrah
2077
$endgroup$
$begingroup$
You're right, thanks. We can use the fact that potential difference between centre of sphere and infinity is $frac3Q8 pi epsilon_0 R$, and so the answer can be difference between my answer in approach two and the potential at the centre of the sphere, that is $$frac3Q8 pi epsilon_0 R - fracQ8 pi epsilon_0 R = fracQ4 pi epsilon_0 R$$ and so we are done.
$endgroup$
– Kushal T.
42 mins ago
add a comment |
$begingroup$
Two cases described are completely different. In first case you find the true potential of the sphere by taking the charge from infinity to the surface of the sphere. In another case you take the charge from the middle of the sphere or the centre of the sphere to the surface of the sphere which is not the potential of the sphere surface. The potential of the sphere surface can be described as the work needed to push a positive charge from infinity to a to the surface or the energy stored to push the charge from the the surface towards the infinity so you can see in your second case you are not calculating the potential of the surface of the sphere. SHORT NOTE:- You can find the potential at any point by finding the difference of potential at that point and any other point whose the potential is zero now at the centre of the the sphere you don't have the potential as 0. See this:http://physics.bu.edu/~duffy/semester2/d06_potential_spheres.html
answered 1 hour ago
Nobody recognizeableNobody recognizeable
657617
$endgroup$
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: "151"
;
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: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
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%2fphysics.stackexchange.com%2fquestions%2f471655%2fpotential-by-assembling-charges%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Approach 2 is wrong. You didn't take into account the corresponding limits for potential. Potential at centre of sphere is not zero!! The expression is V(x)-V(0) instead of V(x).... Find potential at surface by integrating for electric field outside sphere from X to infinity V(infinity)=0. So Then if you wish you can find V(x) by integrating from x=x to any general x=y(
answered 1 hour ago
TojrahTojrah
2077
$endgroup$
$begingroup$
You're right, thanks. We can use the fact that potential difference between centre of sphere and infinity is $frac3Q8 pi epsilon_0 R$, and so the answer can be difference between my answer in approach two and the potential at the centre of the sphere, that is $$frac3Q8 pi epsilon_0 R - fracQ8 pi epsilon_0 R = fracQ4 pi epsilon_0 R$$ and so we are done.
$endgroup$
– Kushal T.
42 mins ago
add a comment |
$begingroup$
Approach 2 is wrong. You didn't take into account the corresponding limits for potential. Potential at centre of sphere is not zero!! The expression is V(x)-V(0) instead of V(x).... Find potential at surface by integrating for electric field outside sphere from X to infinity V(infinity)=0. So Then if you wish you can find V(x) by integrating from x=x to any general x=y(
answered 1 hour ago
TojrahTojrah
2077
$endgroup$
$begingroup$
You're right, thanks. We can use the fact that potential difference between centre of sphere and infinity is $frac3Q8 pi epsilon_0 R$, and so the answer can be difference between my answer in approach two and the potential at the centre of the sphere, that is $$frac3Q8 pi epsilon_0 R - fracQ8 pi epsilon_0 R = fracQ4 pi epsilon_0 R$$ and so we are done.
$endgroup$
– Kushal T.
42 mins ago
add a comment |
$begingroup$
Approach 2 is wrong. You didn't take into account the corresponding limits for potential. Potential at centre of sphere is not zero!! The expression is V(x)-V(0) instead of V(x).... Find potential at surface by integrating for electric field outside sphere from X to infinity V(infinity)=0. So Then if you wish you can find V(x) by integrating from x=x to any general x=y(
answered 1 hour ago
TojrahTojrah
2077
$endgroup$
Approach 2 is wrong. You didn't take into account the corresponding limits for potential. Potential at centre of sphere is not zero!! The expression is V(x)-V(0) instead of V(x).... Find potential at surface by integrating for electric field outside sphere from X to infinity V(infinity)=0. So Then if you wish you can find V(x) by integrating from x=x to any general x=y(
answered 1 hour ago
TojrahTojrah
2077
answered 1 hour ago
TojrahTojrah
2077
answered 1 hour ago
TojrahTojrah
2077
answered 1 hour ago
TojrahTojrah
2077
2077
$begingroup$
You're right, thanks. We can use the fact that potential difference between centre of sphere and infinity is $frac3Q8 pi epsilon_0 R$, and so the answer can be difference between my answer in approach two and the potential at the centre of the sphere, that is $$frac3Q8 pi epsilon_0 R - fracQ8 pi epsilon_0 R = fracQ4 pi epsilon_0 R$$ and so we are done.
$endgroup$
– Kushal T.
42 mins ago
add a comment |
$begingroup$
You're right, thanks. We can use the fact that potential difference between centre of sphere and infinity is $frac3Q8 pi epsilon_0 R$, and so the answer can be difference between my answer in approach two and the potential at the centre of the sphere, that is $$frac3Q8 pi epsilon_0 R - fracQ8 pi epsilon_0 R = fracQ4 pi epsilon_0 R$$ and so we are done.
$endgroup$
– Kushal T.
42 mins ago
$begingroup$
You're right, thanks. We can use the fact that potential difference between centre of sphere and infinity is $frac3Q8 pi epsilon_0 R$, and so the answer can be difference between my answer in approach two and the potential at the centre of the sphere, that is $$frac3Q8 pi epsilon_0 R - fracQ8 pi epsilon_0 R = fracQ4 pi epsilon_0 R$$ and so we are done.
$endgroup$
– Kushal T.
42 mins ago
$begingroup$
You're right, thanks. We can use the fact that potential difference between centre of sphere and infinity is $frac3Q8 pi epsilon_0 R$, and so the answer can be difference between my answer in approach two and the potential at the centre of the sphere, that is $$frac3Q8 pi epsilon_0 R - fracQ8 pi epsilon_0 R = fracQ4 pi epsilon_0 R$$ and so we are done.
$endgroup$
– Kushal T.
42 mins ago
add a comment |
$begingroup$
Two cases described are completely different. In first case you find the true potential of the sphere by taking the charge from infinity to the surface of the sphere. In another case you take the charge from the middle of the sphere or the centre of the sphere to the surface of the sphere which is not the potential of the sphere surface. The potential of the sphere surface can be described as the work needed to push a positive charge from infinity to a to the surface or the energy stored to push the charge from the the surface towards the infinity so you can see in your second case you are not calculating the potential of the surface of the sphere. SHORT NOTE:- You can find the potential at any point by finding the difference of potential at that point and any other point whose the potential is zero now at the centre of the the sphere you don't have the potential as 0. See this:http://physics.bu.edu/~duffy/semester2/d06_potential_spheres.html
answered 1 hour ago
Nobody recognizeableNobody recognizeable
657617
$endgroup$
add a comment |
$begingroup$
Two cases described are completely different. In first case you find the true potential of the sphere by taking the charge from infinity to the surface of the sphere. In another case you take the charge from the middle of the sphere or the centre of the sphere to the surface of the sphere which is not the potential of the sphere surface. The potential of the sphere surface can be described as the work needed to push a positive charge from infinity to a to the surface or the energy stored to push the charge from the the surface towards the infinity so you can see in your second case you are not calculating the potential of the surface of the sphere. SHORT NOTE:- You can find the potential at any point by finding the difference of potential at that point and any other point whose the potential is zero now at the centre of the the sphere you don't have the potential as 0. See this:http://physics.bu.edu/~duffy/semester2/d06_potential_spheres.html
answered 1 hour ago
Nobody recognizeableNobody recognizeable
657617
$endgroup$
add a comment |
$begingroup$
Two cases described are completely different. In first case you find the true potential of the sphere by taking the charge from infinity to the surface of the sphere. In another case you take the charge from the middle of the sphere or the centre of the sphere to the surface of the sphere which is not the potential of the sphere surface. The potential of the sphere surface can be described as the work needed to push a positive charge from infinity to a to the surface or the energy stored to push the charge from the the surface towards the infinity so you can see in your second case you are not calculating the potential of the surface of the sphere. SHORT NOTE:- You can find the potential at any point by finding the difference of potential at that point and any other point whose the potential is zero now at the centre of the the sphere you don't have the potential as 0. See this:http://physics.bu.edu/~duffy/semester2/d06_potential_spheres.html
answered 1 hour ago
Nobody recognizeableNobody recognizeable
657617
$endgroup$
Two cases described are completely different. In first case you find the true potential of the sphere by taking the charge from infinity to the surface of the sphere. In another case you take the charge from the middle of the sphere or the centre of the sphere to the surface of the sphere which is not the potential of the sphere surface. The potential of the sphere surface can be described as the work needed to push a positive charge from infinity to a to the surface or the energy stored to push the charge from the the surface towards the infinity so you can see in your second case you are not calculating the potential of the surface of the sphere. SHORT NOTE:- You can find the potential at any point by finding the difference of potential at that point and any other point whose the potential is zero now at the centre of the the sphere you don't have the potential as 0. See this:http://physics.bu.edu/~duffy/semester2/d06_potential_spheres.html
answered 1 hour ago
Nobody recognizeableNobody recognizeable
657617
edited 48 mins ago
answered 1 hour ago
Nobody recognizeableNobody recognizeable
657617
answered 1 hour ago
Nobody recognizeableNobody recognizeable
657617
answered 1 hour ago
Nobody recognizeableNobody recognizeable
657617
657617
add a comment |
add a comment |
Thanks for contributing an answer to Physics 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%2fphysics.stackexchange.com%2fquestions%2f471655%2fpotential-by-assembling-charges%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
