What if… you had a bowl of electrons?

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My chemistry teacher used to tell us that if you had a soup bowl with only electrons in it, the explosion could make you fly to Pluto. Was he right? Could this happen?










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    My chemistry teacher used to tell us that if you had a soup bowl with only electrons in it, the explosion could make you fly to Pluto. Was he right? Could this happen?










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      My chemistry teacher used to tell us that if you had a soup bowl with only electrons in it, the explosion could make you fly to Pluto. Was he right? Could this happen?










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      My chemistry teacher used to tell us that if you had a soup bowl with only electrons in it, the explosion could make you fly to Pluto. Was he right? Could this happen?







      electromagnetism






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      edited 1 hour ago









      AccidentalFourierTransform

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          2 Answers
          2






          active

          oldest

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          up vote
          5
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          accepted










          The answer would depend how densely the electrons are packed. Let's say we have 1 kg of electrons, meaning we would have about $N = 10^30$ of them. For simplicity, let's approximate by arranging all of these electrons arranged in a spherical shell of radius $r=0.1$ meters. By symmetry, the voltage at the location of each charge would be constant and would have the following value.



          $$V = -frac14pi epsilon_0fracNer$$



          We can thus compute the potential energy $U = frac12Q V$ associated with this configuration.



          $$U = frac12 cdot Ne cdot frac14pi epsilon_0fracNer = fracN^2 e^28 pi epsilon_0 r approx boxed1.15 times 10^33 text Joules$$



          Ignoring air friction, the Earth's escape velocity is $v =11200$ meters / second, which for a $m=2times 10^6$ kg space shuttle, would only require $E = frac12 mv^2 = 1.25 times 10^14$ Joules.



          So, yeah, you'd have way more than plenty sufficient energy.






          share|cite|improve this answer




















          • good job, Trevor. simple and direct.
            – niels nielsen
            1 hour ago










          • Thanks for the answer! So you'd need less than $1mu$g of electrons to get to Pluto :)
            – CuriousStudent
            51 mins ago










          • But will all of that energy be transferred to the object? Also in a spherical shell not all charges are a distance $r$ from all other charges right?
            – Aaron Stevens
            36 mins ago










          • Rigth Aron, but I found this consideration inappropriate for the question. Also, it will suffice anyway
            – user1420303
            25 mins ago










          • @AaronStevens No, but if you integrate over all of the charge it is as if they were. A charged spherical shell acts as if it were all concentrated at the center as far as electric fields outside of the shell are concerned.
            – Trevor Kafka
            18 mins ago

















          up vote
          2
          down vote













          A uniform sphere of radius $r$ and total charge $Q$ has an electric potential energy of $3Q^2/20pi epsilon_0 r$.



          Let's say your bowl is like a sphere of radius 5 cm. If it's all water ( molecular weight 18$m_u$) and $mu= 9/5$ of a mass unit for every electron, then the number of electrons is
          $ N_e = 1000 frac4pi r^33/mu m_u = 5times 10^26$



          If all the nuclei were removed, the electric potential energy would be $7times 10^26$ Joules.



          Pluto is at about 40 au from the Sun, effectively to get there, you have to escape from Earth and escape from the Sun. To do this you would need to launch from the Earth with a carefully directed speed of at least 16 km/s. If your mass is 100 kg, that takes a mere $1.3times 10^10$ J.






          share|cite|improve this answer




















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            5
            down vote



            accepted










            The answer would depend how densely the electrons are packed. Let's say we have 1 kg of electrons, meaning we would have about $N = 10^30$ of them. For simplicity, let's approximate by arranging all of these electrons arranged in a spherical shell of radius $r=0.1$ meters. By symmetry, the voltage at the location of each charge would be constant and would have the following value.



            $$V = -frac14pi epsilon_0fracNer$$



            We can thus compute the potential energy $U = frac12Q V$ associated with this configuration.



            $$U = frac12 cdot Ne cdot frac14pi epsilon_0fracNer = fracN^2 e^28 pi epsilon_0 r approx boxed1.15 times 10^33 text Joules$$



            Ignoring air friction, the Earth's escape velocity is $v =11200$ meters / second, which for a $m=2times 10^6$ kg space shuttle, would only require $E = frac12 mv^2 = 1.25 times 10^14$ Joules.



            So, yeah, you'd have way more than plenty sufficient energy.






            share|cite|improve this answer




















            • good job, Trevor. simple and direct.
              – niels nielsen
              1 hour ago










            • Thanks for the answer! So you'd need less than $1mu$g of electrons to get to Pluto :)
              – CuriousStudent
              51 mins ago










            • But will all of that energy be transferred to the object? Also in a spherical shell not all charges are a distance $r$ from all other charges right?
              – Aaron Stevens
              36 mins ago










            • Rigth Aron, but I found this consideration inappropriate for the question. Also, it will suffice anyway
              – user1420303
              25 mins ago










            • @AaronStevens No, but if you integrate over all of the charge it is as if they were. A charged spherical shell acts as if it were all concentrated at the center as far as electric fields outside of the shell are concerned.
              – Trevor Kafka
              18 mins ago














            up vote
            5
            down vote



            accepted










            The answer would depend how densely the electrons are packed. Let's say we have 1 kg of electrons, meaning we would have about $N = 10^30$ of them. For simplicity, let's approximate by arranging all of these electrons arranged in a spherical shell of radius $r=0.1$ meters. By symmetry, the voltage at the location of each charge would be constant and would have the following value.



            $$V = -frac14pi epsilon_0fracNer$$



            We can thus compute the potential energy $U = frac12Q V$ associated with this configuration.



            $$U = frac12 cdot Ne cdot frac14pi epsilon_0fracNer = fracN^2 e^28 pi epsilon_0 r approx boxed1.15 times 10^33 text Joules$$



            Ignoring air friction, the Earth's escape velocity is $v =11200$ meters / second, which for a $m=2times 10^6$ kg space shuttle, would only require $E = frac12 mv^2 = 1.25 times 10^14$ Joules.



            So, yeah, you'd have way more than plenty sufficient energy.






            share|cite|improve this answer




















            • good job, Trevor. simple and direct.
              – niels nielsen
              1 hour ago










            • Thanks for the answer! So you'd need less than $1mu$g of electrons to get to Pluto :)
              – CuriousStudent
              51 mins ago










            • But will all of that energy be transferred to the object? Also in a spherical shell not all charges are a distance $r$ from all other charges right?
              – Aaron Stevens
              36 mins ago










            • Rigth Aron, but I found this consideration inappropriate for the question. Also, it will suffice anyway
              – user1420303
              25 mins ago










            • @AaronStevens No, but if you integrate over all of the charge it is as if they were. A charged spherical shell acts as if it were all concentrated at the center as far as electric fields outside of the shell are concerned.
              – Trevor Kafka
              18 mins ago












            up vote
            5
            down vote



            accepted







            up vote
            5
            down vote



            accepted






            The answer would depend how densely the electrons are packed. Let's say we have 1 kg of electrons, meaning we would have about $N = 10^30$ of them. For simplicity, let's approximate by arranging all of these electrons arranged in a spherical shell of radius $r=0.1$ meters. By symmetry, the voltage at the location of each charge would be constant and would have the following value.



            $$V = -frac14pi epsilon_0fracNer$$



            We can thus compute the potential energy $U = frac12Q V$ associated with this configuration.



            $$U = frac12 cdot Ne cdot frac14pi epsilon_0fracNer = fracN^2 e^28 pi epsilon_0 r approx boxed1.15 times 10^33 text Joules$$



            Ignoring air friction, the Earth's escape velocity is $v =11200$ meters / second, which for a $m=2times 10^6$ kg space shuttle, would only require $E = frac12 mv^2 = 1.25 times 10^14$ Joules.



            So, yeah, you'd have way more than plenty sufficient energy.






            share|cite|improve this answer












            The answer would depend how densely the electrons are packed. Let's say we have 1 kg of electrons, meaning we would have about $N = 10^30$ of them. For simplicity, let's approximate by arranging all of these electrons arranged in a spherical shell of radius $r=0.1$ meters. By symmetry, the voltage at the location of each charge would be constant and would have the following value.



            $$V = -frac14pi epsilon_0fracNer$$



            We can thus compute the potential energy $U = frac12Q V$ associated with this configuration.



            $$U = frac12 cdot Ne cdot frac14pi epsilon_0fracNer = fracN^2 e^28 pi epsilon_0 r approx boxed1.15 times 10^33 text Joules$$



            Ignoring air friction, the Earth's escape velocity is $v =11200$ meters / second, which for a $m=2times 10^6$ kg space shuttle, would only require $E = frac12 mv^2 = 1.25 times 10^14$ Joules.



            So, yeah, you'd have way more than plenty sufficient energy.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            Trevor Kafka

            490210




            490210











            • good job, Trevor. simple and direct.
              – niels nielsen
              1 hour ago










            • Thanks for the answer! So you'd need less than $1mu$g of electrons to get to Pluto :)
              – CuriousStudent
              51 mins ago










            • But will all of that energy be transferred to the object? Also in a spherical shell not all charges are a distance $r$ from all other charges right?
              – Aaron Stevens
              36 mins ago










            • Rigth Aron, but I found this consideration inappropriate for the question. Also, it will suffice anyway
              – user1420303
              25 mins ago










            • @AaronStevens No, but if you integrate over all of the charge it is as if they were. A charged spherical shell acts as if it were all concentrated at the center as far as electric fields outside of the shell are concerned.
              – Trevor Kafka
              18 mins ago
















            • good job, Trevor. simple and direct.
              – niels nielsen
              1 hour ago










            • Thanks for the answer! So you'd need less than $1mu$g of electrons to get to Pluto :)
              – CuriousStudent
              51 mins ago










            • But will all of that energy be transferred to the object? Also in a spherical shell not all charges are a distance $r$ from all other charges right?
              – Aaron Stevens
              36 mins ago










            • Rigth Aron, but I found this consideration inappropriate for the question. Also, it will suffice anyway
              – user1420303
              25 mins ago










            • @AaronStevens No, but if you integrate over all of the charge it is as if they were. A charged spherical shell acts as if it were all concentrated at the center as far as electric fields outside of the shell are concerned.
              – Trevor Kafka
              18 mins ago















            good job, Trevor. simple and direct.
            – niels nielsen
            1 hour ago




            good job, Trevor. simple and direct.
            – niels nielsen
            1 hour ago












            Thanks for the answer! So you'd need less than $1mu$g of electrons to get to Pluto :)
            – CuriousStudent
            51 mins ago




            Thanks for the answer! So you'd need less than $1mu$g of electrons to get to Pluto :)
            – CuriousStudent
            51 mins ago












            But will all of that energy be transferred to the object? Also in a spherical shell not all charges are a distance $r$ from all other charges right?
            – Aaron Stevens
            36 mins ago




            But will all of that energy be transferred to the object? Also in a spherical shell not all charges are a distance $r$ from all other charges right?
            – Aaron Stevens
            36 mins ago












            Rigth Aron, but I found this consideration inappropriate for the question. Also, it will suffice anyway
            – user1420303
            25 mins ago




            Rigth Aron, but I found this consideration inappropriate for the question. Also, it will suffice anyway
            – user1420303
            25 mins ago












            @AaronStevens No, but if you integrate over all of the charge it is as if they were. A charged spherical shell acts as if it were all concentrated at the center as far as electric fields outside of the shell are concerned.
            – Trevor Kafka
            18 mins ago




            @AaronStevens No, but if you integrate over all of the charge it is as if they were. A charged spherical shell acts as if it were all concentrated at the center as far as electric fields outside of the shell are concerned.
            – Trevor Kafka
            18 mins ago










            up vote
            2
            down vote













            A uniform sphere of radius $r$ and total charge $Q$ has an electric potential energy of $3Q^2/20pi epsilon_0 r$.



            Let's say your bowl is like a sphere of radius 5 cm. If it's all water ( molecular weight 18$m_u$) and $mu= 9/5$ of a mass unit for every electron, then the number of electrons is
            $ N_e = 1000 frac4pi r^33/mu m_u = 5times 10^26$



            If all the nuclei were removed, the electric potential energy would be $7times 10^26$ Joules.



            Pluto is at about 40 au from the Sun, effectively to get there, you have to escape from Earth and escape from the Sun. To do this you would need to launch from the Earth with a carefully directed speed of at least 16 km/s. If your mass is 100 kg, that takes a mere $1.3times 10^10$ J.






            share|cite|improve this answer
























              up vote
              2
              down vote













              A uniform sphere of radius $r$ and total charge $Q$ has an electric potential energy of $3Q^2/20pi epsilon_0 r$.



              Let's say your bowl is like a sphere of radius 5 cm. If it's all water ( molecular weight 18$m_u$) and $mu= 9/5$ of a mass unit for every electron, then the number of electrons is
              $ N_e = 1000 frac4pi r^33/mu m_u = 5times 10^26$



              If all the nuclei were removed, the electric potential energy would be $7times 10^26$ Joules.



              Pluto is at about 40 au from the Sun, effectively to get there, you have to escape from Earth and escape from the Sun. To do this you would need to launch from the Earth with a carefully directed speed of at least 16 km/s. If your mass is 100 kg, that takes a mere $1.3times 10^10$ J.






              share|cite|improve this answer






















                up vote
                2
                down vote










                up vote
                2
                down vote









                A uniform sphere of radius $r$ and total charge $Q$ has an electric potential energy of $3Q^2/20pi epsilon_0 r$.



                Let's say your bowl is like a sphere of radius 5 cm. If it's all water ( molecular weight 18$m_u$) and $mu= 9/5$ of a mass unit for every electron, then the number of electrons is
                $ N_e = 1000 frac4pi r^33/mu m_u = 5times 10^26$



                If all the nuclei were removed, the electric potential energy would be $7times 10^26$ Joules.



                Pluto is at about 40 au from the Sun, effectively to get there, you have to escape from Earth and escape from the Sun. To do this you would need to launch from the Earth with a carefully directed speed of at least 16 km/s. If your mass is 100 kg, that takes a mere $1.3times 10^10$ J.






                share|cite|improve this answer












                A uniform sphere of radius $r$ and total charge $Q$ has an electric potential energy of $3Q^2/20pi epsilon_0 r$.



                Let's say your bowl is like a sphere of radius 5 cm. If it's all water ( molecular weight 18$m_u$) and $mu= 9/5$ of a mass unit for every electron, then the number of electrons is
                $ N_e = 1000 frac4pi r^33/mu m_u = 5times 10^26$



                If all the nuclei were removed, the electric potential energy would be $7times 10^26$ Joules.



                Pluto is at about 40 au from the Sun, effectively to get there, you have to escape from Earth and escape from the Sun. To do this you would need to launch from the Earth with a carefully directed speed of at least 16 km/s. If your mass is 100 kg, that takes a mere $1.3times 10^10$ J.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 45 mins ago









                Rob Jeffries

                65.9k7130223




                65.9k7130223




















                    CuriousStudent is a new contributor. Be nice, and check out our Code of Conduct.









                     

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