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?

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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?










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:



$$rho = frac3Q4 pi R^3$$



$$q = frac43 pi x^3 rho = Q fracx^3R^3$$
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?










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:



    $$rho = frac3Q4 pi R^3$$



    $$q = frac43 pi x^3 rho = Q fracx^3R^3$$
    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?










    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:



      $$rho = frac3Q4 pi R^3$$



      $$q = frac43 pi x^3 rho = Q fracx^3R^3$$
      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?










      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:



      $$rho = frac3Q4 pi R^3$$



      $$q = frac43 pi x^3 rho = Q fracx^3R^3$$
      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited yesterday







      Kushal T.

















      asked yesterday









      Kushal T.Kushal T.

      537




      537




















          3 Answers
          3






          active

          oldest

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          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









          $endgroup$








          • 1




            $begingroup$
            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 $$-frac3Q8 pi epsilon_0 R - ( - fracQ8 pi epsilon_0 R) = boxed-fracQ4 pi epsilon_0 R$$ and so we are done.
            $endgroup$
            – Kushal T.
            yesterday



















          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











          $endgroup$




















            1












            $begingroup$

            The first thing to note is that the electric potential at a point is entirely different to the electric potential energy of an assembly of charges.

            I have assumed that you are finding the potential at a point and you have used two definitions of the zero of electric potential, one at infinity and the other at the centre of the charge distribution.



            Using Gauss's law the graph of electric field strength $E(x)$ against distance from the centre of the charge distribution $x$ looks something like this.



            enter image description here



            The area under the graph $int E,dx$ is related to the change in potential.



            In essence what you have done is found that areas $A$ and $B$ are not the same.



            PS You may well have met a similar graph with $E(r)$ negative and labelled $g(r)$ when discussing the gravitational field due to the earth and the gravitational field strength inside the Earth?






            share|cite|improve this answer









            $endgroup$













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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              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









              $endgroup$








              • 1




                $begingroup$
                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 $$-frac3Q8 pi epsilon_0 R - ( - fracQ8 pi epsilon_0 R) = boxed-fracQ4 pi epsilon_0 R$$ and so we are done.
                $endgroup$
                – Kushal T.
                yesterday
















              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









              $endgroup$








              • 1




                $begingroup$
                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 $$-frac3Q8 pi epsilon_0 R - ( - fracQ8 pi epsilon_0 R) = boxed-fracQ4 pi epsilon_0 R$$ and so we are done.
                $endgroup$
                – Kushal T.
                yesterday














              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 yesterday









              TojrahTojrah

              2207




              2207







              • 1




                $begingroup$
                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 $$-frac3Q8 pi epsilon_0 R - ( - fracQ8 pi epsilon_0 R) = boxed-fracQ4 pi epsilon_0 R$$ and so we are done.
                $endgroup$
                – Kushal T.
                yesterday













              • 1




                $begingroup$
                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 $$-frac3Q8 pi epsilon_0 R - ( - fracQ8 pi epsilon_0 R) = boxed-fracQ4 pi epsilon_0 R$$ and so we are done.
                $endgroup$
                – Kushal T.
                yesterday








              1




              1




              $begingroup$
              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 $$-frac3Q8 pi epsilon_0 R - ( - fracQ8 pi epsilon_0 R) = boxed-fracQ4 pi epsilon_0 R$$ and so we are done.
              $endgroup$
              – Kushal T.
              yesterday





              $begingroup$
              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 $$-frac3Q8 pi epsilon_0 R - ( - fracQ8 pi epsilon_0 R) = boxed-fracQ4 pi epsilon_0 R$$ and so we are done.
              $endgroup$
              – Kushal T.
              yesterday












              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 yesterday

























                  answered yesterday









                  Nobody recognizeableNobody recognizeable

                  657617




                  657617





















                      1












                      $begingroup$

                      The first thing to note is that the electric potential at a point is entirely different to the electric potential energy of an assembly of charges.

                      I have assumed that you are finding the potential at a point and you have used two definitions of the zero of electric potential, one at infinity and the other at the centre of the charge distribution.



                      Using Gauss's law the graph of electric field strength $E(x)$ against distance from the centre of the charge distribution $x$ looks something like this.



                      enter image description here



                      The area under the graph $int E,dx$ is related to the change in potential.



                      In essence what you have done is found that areas $A$ and $B$ are not the same.



                      PS You may well have met a similar graph with $E(r)$ negative and labelled $g(r)$ when discussing the gravitational field due to the earth and the gravitational field strength inside the Earth?






                      share|cite|improve this answer









                      $endgroup$

















                        1












                        $begingroup$

                        The first thing to note is that the electric potential at a point is entirely different to the electric potential energy of an assembly of charges.

                        I have assumed that you are finding the potential at a point and you have used two definitions of the zero of electric potential, one at infinity and the other at the centre of the charge distribution.



                        Using Gauss's law the graph of electric field strength $E(x)$ against distance from the centre of the charge distribution $x$ looks something like this.



                        enter image description here



                        The area under the graph $int E,dx$ is related to the change in potential.



                        In essence what you have done is found that areas $A$ and $B$ are not the same.



                        PS You may well have met a similar graph with $E(r)$ negative and labelled $g(r)$ when discussing the gravitational field due to the earth and the gravitational field strength inside the Earth?






                        share|cite|improve this answer









                        $endgroup$















                          1












                          1








                          1





                          $begingroup$

                          The first thing to note is that the electric potential at a point is entirely different to the electric potential energy of an assembly of charges.

                          I have assumed that you are finding the potential at a point and you have used two definitions of the zero of electric potential, one at infinity and the other at the centre of the charge distribution.



                          Using Gauss's law the graph of electric field strength $E(x)$ against distance from the centre of the charge distribution $x$ looks something like this.



                          enter image description here



                          The area under the graph $int E,dx$ is related to the change in potential.



                          In essence what you have done is found that areas $A$ and $B$ are not the same.



                          PS You may well have met a similar graph with $E(r)$ negative and labelled $g(r)$ when discussing the gravitational field due to the earth and the gravitational field strength inside the Earth?






                          share|cite|improve this answer









                          $endgroup$



                          The first thing to note is that the electric potential at a point is entirely different to the electric potential energy of an assembly of charges.

                          I have assumed that you are finding the potential at a point and you have used two definitions of the zero of electric potential, one at infinity and the other at the centre of the charge distribution.



                          Using Gauss's law the graph of electric field strength $E(x)$ against distance from the centre of the charge distribution $x$ looks something like this.



                          enter image description here



                          The area under the graph $int E,dx$ is related to the change in potential.



                          In essence what you have done is found that areas $A$ and $B$ are not the same.



                          PS You may well have met a similar graph with $E(r)$ negative and labelled $g(r)$ when discussing the gravitational field due to the earth and the gravitational field strength inside the Earth?







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered yesterday









                          FarcherFarcher

                          52.1k340109




                          52.1k340109



























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