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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?












1












$begingroup$


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











$endgroup$

















    1












    $begingroup$


    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











    $endgroup$















      1












      1








      1





      $begingroup$


      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 5 secs ago







      K. T

















      asked 1 hour ago









      K. TK. T

      364




      364






















          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.
            $endgroup$
            – K. T
            16 mins ago



















          1












          $begingroup$

          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











          $endgroup$














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            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.
              $endgroup$
              – K. T
              16 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.
              $endgroup$
              – K. T
              16 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 47 mins 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.
              $endgroup$
              – K. T
              16 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.
              $endgroup$
              – K. T
              16 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.
            $endgroup$
            – K. T
            16 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.
            $endgroup$
            – K. T
            16 mins ago











            1












            $begingroup$

            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












              $begingroup$

              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












                1








                1





                $begingroup$

                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 21 mins ago

























                answered 43 mins ago









                Nobody recognizeableNobody recognizeable

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