Potential by Assembling Charges












2












$begingroup$


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:



$$rho = frac{3Q}{4 pi R^{3}}$$



$$q = frac{4}{3} pi x^{3} rho = Q frac{x^{3}}{R^3}$$
Potential at the surface would be $$V = frac{q}{4 pi epsilon_0 x} = frac{Q x^{2}}{4 pi epsilon_0 R^{3}}$$



Approach 2:
$$rho = frac{3Q}{4 pi R^{3}}$$
$$q = frac{4}{3} pi x^{3} rho = Q frac{x^{3}}{R^3}$$
$$E = frac{Q x}{4 pi epsilon_0 R^{3}}$$ (From Gauss' Law)



Potential at the surface would be $$V = -int{vec{E} cdot vec{dx}} = -frac{Q}{4 pi epsilon_0 R^{3}} int_{0}^{x}{xdx} = -frac{Q x^{2}}{8 pi epsilon_0 R^{3}}$$



Why is the answer different in both the cases?






electrostatics potential







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    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:



    $$rho = frac{3Q}{4 pi R^{3}}$$



    $$q = frac{4}{3} pi x^{3} rho = Q frac{x^{3}}{R^3}$$
    Potential at the surface would be $$V = frac{q}{4 pi epsilon_0 x} = frac{Q x^{2}}{4 pi epsilon_0 R^{3}}$$



    Approach 2:
    $$rho = frac{3Q}{4 pi R^{3}}$$
    $$q = frac{4}{3} pi x^{3} rho = Q frac{x^{3}}{R^3}$$
    $$E = frac{Q x}{4 pi epsilon_0 R^{3}}$$ (From Gauss' Law)



    Potential at the surface would be $$V = -int{vec{E} cdot vec{dx}} = -frac{Q}{4 pi epsilon_0 R^{3}} int_{0}^{x}{xdx} = -frac{Q x^{2}}{8 pi epsilon_0 R^{3}}$$



    Why is the answer different in both the cases?






    electrostatics potential







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      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:



      $$rho = frac{3Q}{4 pi R^{3}}$$



      $$q = frac{4}{3} pi x^{3} rho = Q frac{x^{3}}{R^3}$$
      Potential at the surface would be $$V = frac{q}{4 pi epsilon_0 x} = frac{Q x^{2}}{4 pi epsilon_0 R^{3}}$$



      Approach 2:
      $$rho = frac{3Q}{4 pi R^{3}}$$
      $$q = frac{4}{3} pi x^{3} rho = Q frac{x^{3}}{R^3}$$
      $$E = frac{Q x}{4 pi epsilon_0 R^{3}}$$ (From Gauss' Law)



      Potential at the surface would be $$V = -int{vec{E} cdot vec{dx}} = -frac{Q}{4 pi epsilon_0 R^{3}} int_{0}^{x}{xdx} = -frac{Q x^{2}}{8 pi epsilon_0 R^{3}}$$



      Why is the answer different in both the cases?






      electrostatics potential







      share|cite|improve this question











      $endgroup$




      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:



      $$rho = frac{3Q}{4 pi R^{3}}$$



      $$q = frac{4}{3} pi x^{3} rho = Q frac{x^{3}}{R^3}$$
      Potential at the surface would be $$V = frac{q}{4 pi epsilon_0 x} = frac{Q x^{2}}{4 pi epsilon_0 R^{3}}$$



      Approach 2:
      $$rho = frac{3Q}{4 pi R^{3}}$$
      $$q = frac{4}{3} pi x^{3} rho = Q frac{x^{3}}{R^3}$$
      $$E = frac{Q x}{4 pi epsilon_0 R^{3}}$$ (From Gauss' Law)



      Potential at the surface would be $$V = -int{vec{E} cdot vec{dx}} = -frac{Q}{4 pi epsilon_0 R^{3}} int_{0}^{x}{xdx} = -frac{Q x^{2}}{8 pi epsilon_0 R^{3}}$$



      Why is the answer different in both the cases?






      electrostatics potential




      electrostatics potential






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 26 mins ago







      Kushal T.

















      asked 1 hour ago









      Kushal T.Kushal T.

      416




      416






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          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(






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            You're right, thanks. We can use the fact that potential difference between centre of sphere and infinity is $frac{3Q}{8 pi epsilon_0 R}$, and so the answer can be difference between my answer in approach two and the potential at the centre of the sphere, that is $$frac{3Q}{8 pi epsilon_0 R} - frac{Q}{8 pi epsilon_0 R} = frac{Q}{4 pi epsilon_0 R}$$ and so we are done.
            $endgroup$
            – Kushal T.
            42 mins ago



















          2












          $begingroup$

          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






          share|cite|improve this answer











          $endgroup$














            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
            });


            }
            });














            draft saved

            draft discarded


















            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









            2












            $begingroup$

            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(






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You're right, thanks. We can use the fact that potential difference between centre of sphere and infinity is $frac{3Q}{8 pi epsilon_0 R}$, and so the answer can be difference between my answer in approach two and the potential at the centre of the sphere, that is $$frac{3Q}{8 pi epsilon_0 R} - frac{Q}{8 pi epsilon_0 R} = frac{Q}{4 pi epsilon_0 R}$$ and so we are done.
              $endgroup$
              – Kushal T.
              42 mins ago
















            2












            $begingroup$

            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(






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              You're right, thanks. We can use the fact that potential difference between centre of sphere and infinity is $frac{3Q}{8 pi epsilon_0 R}$, and so the answer can be difference between my answer in approach two and the potential at the centre of the sphere, that is $$frac{3Q}{8 pi epsilon_0 R} - frac{Q}{8 pi epsilon_0 R} = frac{Q}{4 pi epsilon_0 R}$$ and so we are done.
              $endgroup$
              – Kushal T.
              42 mins ago














            2












            2








            2





            $begingroup$

            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(






            share|cite|improve this answer









            $endgroup$



            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(







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            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 $frac{3Q}{8 pi epsilon_0 R}$, and so the answer can be difference between my answer in approach two and the potential at the centre of the sphere, that is $$frac{3Q}{8 pi epsilon_0 R} - frac{Q}{8 pi epsilon_0 R} = frac{Q}{4 pi epsilon_0 R}$$ 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 $frac{3Q}{8 pi epsilon_0 R}$, and so the answer can be difference between my answer in approach two and the potential at the centre of the sphere, that is $$frac{3Q}{8 pi epsilon_0 R} - frac{Q}{8 pi epsilon_0 R} = frac{Q}{4 pi epsilon_0 R}$$ 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 $frac{3Q}{8 pi epsilon_0 R}$, and so the answer can be difference between my answer in approach two and the potential at the centre of the sphere, that is $$frac{3Q}{8 pi epsilon_0 R} - frac{Q}{8 pi epsilon_0 R} = frac{Q}{4 pi epsilon_0 R}$$ 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 $frac{3Q}{8 pi epsilon_0 R}$, and so the answer can be difference between my answer in approach two and the potential at the centre of the sphere, that is $$frac{3Q}{8 pi epsilon_0 R} - frac{Q}{8 pi epsilon_0 R} = frac{Q}{4 pi epsilon_0 R}$$ and so we are done.
            $endgroup$
            – Kushal T.
            42 mins ago











            2












            $begingroup$

            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






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              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






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                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






                share|cite|improve this answer











                $endgroup$



                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







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 48 mins ago

























                answered 1 hour ago









                Nobody recognizeableNobody recognizeable

                657617




                657617






























                    draft saved

                    draft discarded




















































                    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.




                    draft saved


                    draft discarded














                    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





















































                    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







                    -

                    Popular posts from this blog

                    Сан-Квентин

                    Image
                    Сан-Квентин Местоположение Сан-Квентин, Калифорния Координаты 37°56′13″ с. ш. 122°29′20″ з. д. H G Я O L Текущий статус Действует Режим безопасности минимальный и максимальный Количество мест 3 302 Открытие Июнь 1852 Находится в ведомстве California Department of Corrections and Rehabilitation Начальник Mike Martel, начальник  Сан-Квентин на Викискладе Сан-Квентин (англ.  San Quentin ) — тюрьма, располагающаяся на 432 акрах (около 1,7 км²), на мысе Сан-Квентин, в округе Марин, штат Калифорния, США. Тюрьма штата Калифорния Сан-Квентин была открыта в июле 1852 года и является старейшей в штате. Она была построена заключёнными, которые проживали в течение строительства на тюремном судне Вабан. В Сан-Квентине отбывали наказание и мужчины, и женщины до 1934 года, когда построили женскую тюрьму в Техачапи и туда перевели всех заключённых женщин. В Сан-Квентине приводятся в исполнение смертные казни, для этого имеется газовая камера, но...
                    Read more

                    8-я гвардейская общевойсковая армия

                    Image
                    8-я гвардейская общевойсковая ордена Ленина армия Награды: Войска: РККА → Советская армия → СВ России Род войск: Формирование: 1943 Расформирование (преобразование): 1992 Предшественник: 62-я армия (1942—43) → 8-я гвардейская общевойсковая армия (1943—92) → 8-й гвардейский армейский корпус (1992—98) Боевой путь Изюм-Барвенковская операция Донбасская операция Запорожская операция Битва за Днепр Днепропетровская операция Березнеговато-Снигирёвская операция Одесская операция Люблин-Брестская операция Варшавско-Познанская операция Восточно-Померанская операция Берлинская наступательная операция У этого термина существуют и другие значения, см. 8-я армия. Фотография Гвардейского Красного Знамени (лицевая сторона) 8-й гвардейской армии , образца 1943 года. Фотография Гвардейского Красного Знамени (оборотная сторона) 8-й гвардейской армии , образца 1943 года. 8-я гвардейская общевойсковая ордена Ленина армия  — гвардейское фор...
                    Read more

                    Алькесар

                    Image
                    Муниципалитет Алькесар Alquézar Герб 42°10′16″ с. ш. 0°01′27″ в. д. H G Я O L Страна Испания   Испания Автономное сообщество Арагон Провинция Уэска Комарка Сомонтано-де-Барбастро Алькальд Хосе Марияно Альтемир Ласкорс (ИСРП) История и география Площадь 32,36 км² Высота НУМ 660 м Часовой пояс UTC+1, летом UTC+2 Население Население 313 человек ( 2010 ) Плотность 9,92 чел./км² Этнохороним alquezarano/-a Цифровые идентификаторы Телефонный код +34 974 Почтовые индексы 22112 Автомобильный код HU alquezar.org   (исп.) Показать/скрыть карты Алькесар Алькесар Алькесар  Аудио, фото и видео на Викискладе Алькесар (исп.  Alquézar ) — муниципалитет в Испании, входит в провинцию Уэска, в составе автономного сообщества Арагон. Муниципалитет находится в составе района (комарки) Сомонтано-де-Барбастро. Занимает площадь 32,36 км². Население — 321 челове...
                    Read more