Charged enclosed by the sphere












4












$begingroup$


I'm reviewing the book "Conquering the Physics GRE" for my upcoming Physics GRE. I came across this problem which I'm having trouble with understanding. In particular, I understand the solution that the author provides but I don't understand what is wrong with my approach.



Q. The Electric field inside a sphere of radius $R$ is given by $E = E_0 z^2 hat{textbf{z}}$. What is the total charge of the sphere ?



The authors approach involving taking the divergence of the electric field to get the charge density and then integrating the density over the volume of the sphere to get charged enclosed, which in their case turns out to be $0$.



But we can also just use a concentric sphere of radius $r$ ($0 < r le R$) as a Gaussian surface and just use the integral form of Maxwell's equation to calculate the charge enclosed.



$$ oint limits_{S} vec{E} cdot dvec{S} = frac{Q_{enc}}{epsilon_0} $$



Since the area vector points in the radial direction, if we assume it makes an angle $theta$ with the Electric Field vector, and given $z = r cos(theta)$, we have



$$ Q_{enc} = epsilon_0 int limits_{0}^{pi} int limits_{0}^{2pi} E_0 r^2 cos^2(theta) r^2 sin(theta) dtheta dphi $$



$$ Q_{enc} = frac{4 pi epsilon_0 E_0}{3} r^4 $$



If we want the charge enclosed by the sphere, we just set $r = R$, so we get



$$ Q = frac{4 pi epsilon_0 E_0}{3} R^4 $$



which isn't zero.



I'm having trouble figuring out where I'm going wrong. Any suggestions appreciated.










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timoneo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    4












    $begingroup$


    I'm reviewing the book "Conquering the Physics GRE" for my upcoming Physics GRE. I came across this problem which I'm having trouble with understanding. In particular, I understand the solution that the author provides but I don't understand what is wrong with my approach.



    Q. The Electric field inside a sphere of radius $R$ is given by $E = E_0 z^2 hat{textbf{z}}$. What is the total charge of the sphere ?



    The authors approach involving taking the divergence of the electric field to get the charge density and then integrating the density over the volume of the sphere to get charged enclosed, which in their case turns out to be $0$.



    But we can also just use a concentric sphere of radius $r$ ($0 < r le R$) as a Gaussian surface and just use the integral form of Maxwell's equation to calculate the charge enclosed.



    $$ oint limits_{S} vec{E} cdot dvec{S} = frac{Q_{enc}}{epsilon_0} $$



    Since the area vector points in the radial direction, if we assume it makes an angle $theta$ with the Electric Field vector, and given $z = r cos(theta)$, we have



    $$ Q_{enc} = epsilon_0 int limits_{0}^{pi} int limits_{0}^{2pi} E_0 r^2 cos^2(theta) r^2 sin(theta) dtheta dphi $$



    $$ Q_{enc} = frac{4 pi epsilon_0 E_0}{3} r^4 $$



    If we want the charge enclosed by the sphere, we just set $r = R$, so we get



    $$ Q = frac{4 pi epsilon_0 E_0}{3} R^4 $$



    which isn't zero.



    I'm having trouble figuring out where I'm going wrong. Any suggestions appreciated.










    share|cite|improve this question







    New contributor




    timoneo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      4












      4








      4


      3



      $begingroup$


      I'm reviewing the book "Conquering the Physics GRE" for my upcoming Physics GRE. I came across this problem which I'm having trouble with understanding. In particular, I understand the solution that the author provides but I don't understand what is wrong with my approach.



      Q. The Electric field inside a sphere of radius $R$ is given by $E = E_0 z^2 hat{textbf{z}}$. What is the total charge of the sphere ?



      The authors approach involving taking the divergence of the electric field to get the charge density and then integrating the density over the volume of the sphere to get charged enclosed, which in their case turns out to be $0$.



      But we can also just use a concentric sphere of radius $r$ ($0 < r le R$) as a Gaussian surface and just use the integral form of Maxwell's equation to calculate the charge enclosed.



      $$ oint limits_{S} vec{E} cdot dvec{S} = frac{Q_{enc}}{epsilon_0} $$



      Since the area vector points in the radial direction, if we assume it makes an angle $theta$ with the Electric Field vector, and given $z = r cos(theta)$, we have



      $$ Q_{enc} = epsilon_0 int limits_{0}^{pi} int limits_{0}^{2pi} E_0 r^2 cos^2(theta) r^2 sin(theta) dtheta dphi $$



      $$ Q_{enc} = frac{4 pi epsilon_0 E_0}{3} r^4 $$



      If we want the charge enclosed by the sphere, we just set $r = R$, so we get



      $$ Q = frac{4 pi epsilon_0 E_0}{3} R^4 $$



      which isn't zero.



      I'm having trouble figuring out where I'm going wrong. Any suggestions appreciated.










      share|cite|improve this question







      New contributor




      timoneo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I'm reviewing the book "Conquering the Physics GRE" for my upcoming Physics GRE. I came across this problem which I'm having trouble with understanding. In particular, I understand the solution that the author provides but I don't understand what is wrong with my approach.



      Q. The Electric field inside a sphere of radius $R$ is given by $E = E_0 z^2 hat{textbf{z}}$. What is the total charge of the sphere ?



      The authors approach involving taking the divergence of the electric field to get the charge density and then integrating the density over the volume of the sphere to get charged enclosed, which in their case turns out to be $0$.



      But we can also just use a concentric sphere of radius $r$ ($0 < r le R$) as a Gaussian surface and just use the integral form of Maxwell's equation to calculate the charge enclosed.



      $$ oint limits_{S} vec{E} cdot dvec{S} = frac{Q_{enc}}{epsilon_0} $$



      Since the area vector points in the radial direction, if we assume it makes an angle $theta$ with the Electric Field vector, and given $z = r cos(theta)$, we have



      $$ Q_{enc} = epsilon_0 int limits_{0}^{pi} int limits_{0}^{2pi} E_0 r^2 cos^2(theta) r^2 sin(theta) dtheta dphi $$



      $$ Q_{enc} = frac{4 pi epsilon_0 E_0}{3} r^4 $$



      If we want the charge enclosed by the sphere, we just set $r = R$, so we get



      $$ Q = frac{4 pi epsilon_0 E_0}{3} R^4 $$



      which isn't zero.



      I'm having trouble figuring out where I'm going wrong. Any suggestions appreciated.







      electrostatics gauss-law maxwell-equations






      share|cite|improve this question







      New contributor




      timoneo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      timoneo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









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      share|cite|improve this question






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      asked 5 hours ago









      timoneotimoneo

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          2 Answers
          2






          active

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          4












          $begingroup$

          I think you forgot to account for $mathbf{hat{z}}$



          Let $mathbf{hat{r}}$ be the normal to the surface of our sphere. If you take the route of integrating the electric field over the surface of the sphere that contains the charge, then you will be evaluating the following quantity.



          $z^2 mathbf{hat{z}}.mathbf{hat{r}}=z^2 mathbf{hat{z}}.mathbf{r}/r=z^2, z/r=z^3/r$



          So you will be integrating $z^3$ over the surface of the sphere centered on the origin. Since $z^3$ is an odd function the integral will vanish.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks! Indeed I forgot the $hat{z}$. FYI, I marked your answer as correct, but I don't have enough reputation to publicly upvote you, so just thanking you via this comment.
            $endgroup$
            – timoneo
            4 hours ago










          • $begingroup$
            I thought this question was fascinating. I don't understand where the minus sign appears to make the function odd. Shouldn't ^z and ^r point in the same direction over the entire surface of the sphere, giving a positive dot product over the entire surface?
            $endgroup$
            – lamplamp
            3 hours ago








          • 1




            $begingroup$
            @timoneo: good question. Glad I could help
            $endgroup$
            – Cryo
            3 hours ago








          • 1




            $begingroup$
            @lamplamp: $mathbf{hat{z}}$ does point in the same direction at all points, but $mathbf{hat{r}}$ does not since it is normal and points out of the sphere. If this sphere was Earth, $mathbf{hat{r}}$ would point in the direction of the rocket taking off the Earth and flying to space, so on north pole $mathbf{hat{r}}$ points "up", whilst on south pole it points "down".
            $endgroup$
            – Cryo
            3 hours ago












          • $begingroup$
            Thanks, when I read the question, I assumed that z was referencing a radial coordinate as a dummy variable to distinguish from r. From the thread here, I now believe that z was chosen to as standard Cartesian, which makes perfect sense about the function being odd.
            $endgroup$
            – lamplamp
            3 hours ago



















          2












          $begingroup$

          There is also a nice geometrical argument for this. Since the field is $vec E=z^2hat z$, the fields lines always point along $+hat z$ and the magnitude of the field does not depend on the position in the $xy$-plane. As a result, every field line that enters the sphere must also exit the sphere, so the net flux must be $0$.



          enter image description here






          share|cite|improve this answer











          $endgroup$













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            2 Answers
            2






            active

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            2 Answers
            2






            active

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            active

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            votes






            active

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            4












            $begingroup$

            I think you forgot to account for $mathbf{hat{z}}$



            Let $mathbf{hat{r}}$ be the normal to the surface of our sphere. If you take the route of integrating the electric field over the surface of the sphere that contains the charge, then you will be evaluating the following quantity.



            $z^2 mathbf{hat{z}}.mathbf{hat{r}}=z^2 mathbf{hat{z}}.mathbf{r}/r=z^2, z/r=z^3/r$



            So you will be integrating $z^3$ over the surface of the sphere centered on the origin. Since $z^3$ is an odd function the integral will vanish.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks! Indeed I forgot the $hat{z}$. FYI, I marked your answer as correct, but I don't have enough reputation to publicly upvote you, so just thanking you via this comment.
              $endgroup$
              – timoneo
              4 hours ago










            • $begingroup$
              I thought this question was fascinating. I don't understand where the minus sign appears to make the function odd. Shouldn't ^z and ^r point in the same direction over the entire surface of the sphere, giving a positive dot product over the entire surface?
              $endgroup$
              – lamplamp
              3 hours ago








            • 1




              $begingroup$
              @timoneo: good question. Glad I could help
              $endgroup$
              – Cryo
              3 hours ago








            • 1




              $begingroup$
              @lamplamp: $mathbf{hat{z}}$ does point in the same direction at all points, but $mathbf{hat{r}}$ does not since it is normal and points out of the sphere. If this sphere was Earth, $mathbf{hat{r}}$ would point in the direction of the rocket taking off the Earth and flying to space, so on north pole $mathbf{hat{r}}$ points "up", whilst on south pole it points "down".
              $endgroup$
              – Cryo
              3 hours ago












            • $begingroup$
              Thanks, when I read the question, I assumed that z was referencing a radial coordinate as a dummy variable to distinguish from r. From the thread here, I now believe that z was chosen to as standard Cartesian, which makes perfect sense about the function being odd.
              $endgroup$
              – lamplamp
              3 hours ago
















            4












            $begingroup$

            I think you forgot to account for $mathbf{hat{z}}$



            Let $mathbf{hat{r}}$ be the normal to the surface of our sphere. If you take the route of integrating the electric field over the surface of the sphere that contains the charge, then you will be evaluating the following quantity.



            $z^2 mathbf{hat{z}}.mathbf{hat{r}}=z^2 mathbf{hat{z}}.mathbf{r}/r=z^2, z/r=z^3/r$



            So you will be integrating $z^3$ over the surface of the sphere centered on the origin. Since $z^3$ is an odd function the integral will vanish.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thanks! Indeed I forgot the $hat{z}$. FYI, I marked your answer as correct, but I don't have enough reputation to publicly upvote you, so just thanking you via this comment.
              $endgroup$
              – timoneo
              4 hours ago










            • $begingroup$
              I thought this question was fascinating. I don't understand where the minus sign appears to make the function odd. Shouldn't ^z and ^r point in the same direction over the entire surface of the sphere, giving a positive dot product over the entire surface?
              $endgroup$
              – lamplamp
              3 hours ago








            • 1




              $begingroup$
              @timoneo: good question. Glad I could help
              $endgroup$
              – Cryo
              3 hours ago








            • 1




              $begingroup$
              @lamplamp: $mathbf{hat{z}}$ does point in the same direction at all points, but $mathbf{hat{r}}$ does not since it is normal and points out of the sphere. If this sphere was Earth, $mathbf{hat{r}}$ would point in the direction of the rocket taking off the Earth and flying to space, so on north pole $mathbf{hat{r}}$ points "up", whilst on south pole it points "down".
              $endgroup$
              – Cryo
              3 hours ago












            • $begingroup$
              Thanks, when I read the question, I assumed that z was referencing a radial coordinate as a dummy variable to distinguish from r. From the thread here, I now believe that z was chosen to as standard Cartesian, which makes perfect sense about the function being odd.
              $endgroup$
              – lamplamp
              3 hours ago














            4












            4








            4





            $begingroup$

            I think you forgot to account for $mathbf{hat{z}}$



            Let $mathbf{hat{r}}$ be the normal to the surface of our sphere. If you take the route of integrating the electric field over the surface of the sphere that contains the charge, then you will be evaluating the following quantity.



            $z^2 mathbf{hat{z}}.mathbf{hat{r}}=z^2 mathbf{hat{z}}.mathbf{r}/r=z^2, z/r=z^3/r$



            So you will be integrating $z^3$ over the surface of the sphere centered on the origin. Since $z^3$ is an odd function the integral will vanish.






            share|cite|improve this answer









            $endgroup$



            I think you forgot to account for $mathbf{hat{z}}$



            Let $mathbf{hat{r}}$ be the normal to the surface of our sphere. If you take the route of integrating the electric field over the surface of the sphere that contains the charge, then you will be evaluating the following quantity.



            $z^2 mathbf{hat{z}}.mathbf{hat{r}}=z^2 mathbf{hat{z}}.mathbf{r}/r=z^2, z/r=z^3/r$



            So you will be integrating $z^3$ over the surface of the sphere centered on the origin. Since $z^3$ is an odd function the integral will vanish.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 4 hours ago









            CryoCryo

            39815




            39815












            • $begingroup$
              Thanks! Indeed I forgot the $hat{z}$. FYI, I marked your answer as correct, but I don't have enough reputation to publicly upvote you, so just thanking you via this comment.
              $endgroup$
              – timoneo
              4 hours ago










            • $begingroup$
              I thought this question was fascinating. I don't understand where the minus sign appears to make the function odd. Shouldn't ^z and ^r point in the same direction over the entire surface of the sphere, giving a positive dot product over the entire surface?
              $endgroup$
              – lamplamp
              3 hours ago








            • 1




              $begingroup$
              @timoneo: good question. Glad I could help
              $endgroup$
              – Cryo
              3 hours ago








            • 1




              $begingroup$
              @lamplamp: $mathbf{hat{z}}$ does point in the same direction at all points, but $mathbf{hat{r}}$ does not since it is normal and points out of the sphere. If this sphere was Earth, $mathbf{hat{r}}$ would point in the direction of the rocket taking off the Earth and flying to space, so on north pole $mathbf{hat{r}}$ points "up", whilst on south pole it points "down".
              $endgroup$
              – Cryo
              3 hours ago












            • $begingroup$
              Thanks, when I read the question, I assumed that z was referencing a radial coordinate as a dummy variable to distinguish from r. From the thread here, I now believe that z was chosen to as standard Cartesian, which makes perfect sense about the function being odd.
              $endgroup$
              – lamplamp
              3 hours ago


















            • $begingroup$
              Thanks! Indeed I forgot the $hat{z}$. FYI, I marked your answer as correct, but I don't have enough reputation to publicly upvote you, so just thanking you via this comment.
              $endgroup$
              – timoneo
              4 hours ago










            • $begingroup$
              I thought this question was fascinating. I don't understand where the minus sign appears to make the function odd. Shouldn't ^z and ^r point in the same direction over the entire surface of the sphere, giving a positive dot product over the entire surface?
              $endgroup$
              – lamplamp
              3 hours ago








            • 1




              $begingroup$
              @timoneo: good question. Glad I could help
              $endgroup$
              – Cryo
              3 hours ago








            • 1




              $begingroup$
              @lamplamp: $mathbf{hat{z}}$ does point in the same direction at all points, but $mathbf{hat{r}}$ does not since it is normal and points out of the sphere. If this sphere was Earth, $mathbf{hat{r}}$ would point in the direction of the rocket taking off the Earth and flying to space, so on north pole $mathbf{hat{r}}$ points "up", whilst on south pole it points "down".
              $endgroup$
              – Cryo
              3 hours ago












            • $begingroup$
              Thanks, when I read the question, I assumed that z was referencing a radial coordinate as a dummy variable to distinguish from r. From the thread here, I now believe that z was chosen to as standard Cartesian, which makes perfect sense about the function being odd.
              $endgroup$
              – lamplamp
              3 hours ago
















            $begingroup$
            Thanks! Indeed I forgot the $hat{z}$. FYI, I marked your answer as correct, but I don't have enough reputation to publicly upvote you, so just thanking you via this comment.
            $endgroup$
            – timoneo
            4 hours ago




            $begingroup$
            Thanks! Indeed I forgot the $hat{z}$. FYI, I marked your answer as correct, but I don't have enough reputation to publicly upvote you, so just thanking you via this comment.
            $endgroup$
            – timoneo
            4 hours ago












            $begingroup$
            I thought this question was fascinating. I don't understand where the minus sign appears to make the function odd. Shouldn't ^z and ^r point in the same direction over the entire surface of the sphere, giving a positive dot product over the entire surface?
            $endgroup$
            – lamplamp
            3 hours ago






            $begingroup$
            I thought this question was fascinating. I don't understand where the minus sign appears to make the function odd. Shouldn't ^z and ^r point in the same direction over the entire surface of the sphere, giving a positive dot product over the entire surface?
            $endgroup$
            – lamplamp
            3 hours ago






            1




            1




            $begingroup$
            @timoneo: good question. Glad I could help
            $endgroup$
            – Cryo
            3 hours ago






            $begingroup$
            @timoneo: good question. Glad I could help
            $endgroup$
            – Cryo
            3 hours ago






            1




            1




            $begingroup$
            @lamplamp: $mathbf{hat{z}}$ does point in the same direction at all points, but $mathbf{hat{r}}$ does not since it is normal and points out of the sphere. If this sphere was Earth, $mathbf{hat{r}}$ would point in the direction of the rocket taking off the Earth and flying to space, so on north pole $mathbf{hat{r}}$ points "up", whilst on south pole it points "down".
            $endgroup$
            – Cryo
            3 hours ago






            $begingroup$
            @lamplamp: $mathbf{hat{z}}$ does point in the same direction at all points, but $mathbf{hat{r}}$ does not since it is normal and points out of the sphere. If this sphere was Earth, $mathbf{hat{r}}$ would point in the direction of the rocket taking off the Earth and flying to space, so on north pole $mathbf{hat{r}}$ points "up", whilst on south pole it points "down".
            $endgroup$
            – Cryo
            3 hours ago














            $begingroup$
            Thanks, when I read the question, I assumed that z was referencing a radial coordinate as a dummy variable to distinguish from r. From the thread here, I now believe that z was chosen to as standard Cartesian, which makes perfect sense about the function being odd.
            $endgroup$
            – lamplamp
            3 hours ago




            $begingroup$
            Thanks, when I read the question, I assumed that z was referencing a radial coordinate as a dummy variable to distinguish from r. From the thread here, I now believe that z was chosen to as standard Cartesian, which makes perfect sense about the function being odd.
            $endgroup$
            – lamplamp
            3 hours ago











            2












            $begingroup$

            There is also a nice geometrical argument for this. Since the field is $vec E=z^2hat z$, the fields lines always point along $+hat z$ and the magnitude of the field does not depend on the position in the $xy$-plane. As a result, every field line that enters the sphere must also exit the sphere, so the net flux must be $0$.



            enter image description here






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              There is also a nice geometrical argument for this. Since the field is $vec E=z^2hat z$, the fields lines always point along $+hat z$ and the magnitude of the field does not depend on the position in the $xy$-plane. As a result, every field line that enters the sphere must also exit the sphere, so the net flux must be $0$.



              enter image description here






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                There is also a nice geometrical argument for this. Since the field is $vec E=z^2hat z$, the fields lines always point along $+hat z$ and the magnitude of the field does not depend on the position in the $xy$-plane. As a result, every field line that enters the sphere must also exit the sphere, so the net flux must be $0$.



                enter image description here






                share|cite|improve this answer











                $endgroup$



                There is also a nice geometrical argument for this. Since the field is $vec E=z^2hat z$, the fields lines always point along $+hat z$ and the magnitude of the field does not depend on the position in the $xy$-plane. As a result, every field line that enters the sphere must also exit the sphere, so the net flux must be $0$.



                enter image description here







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 hours ago

























                answered 2 hours ago









                ZeroTheHeroZeroTheHero

                20.5k53260




                20.5k53260






















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