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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?
electromagnetism
edited 1 hour ago
AccidentalFourierTransform
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CuriousStudent
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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
edited 1 hour ago
AccidentalFourierTransform
23.9k1365119
asked 1 hour ago
CuriousStudent
112
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CuriousStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
1
down vote
favorite
up vote
1
down vote
favorite
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
edited 1 hour ago
AccidentalFourierTransform
23.9k1365119
asked 1 hour ago
CuriousStudent
112
New contributor
CuriousStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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
electromagnetism
edited 1 hour ago
AccidentalFourierTransform
23.9k1365119
asked 1 hour ago
CuriousStudent
112
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CuriousStudent is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 1 hour ago
AccidentalFourierTransform
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asked 1 hour ago
CuriousStudent
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edited 1 hour ago
AccidentalFourierTransform
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edited 1 hour ago
AccidentalFourierTransform
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edited 1 hour ago
AccidentalFourierTransform
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CuriousStudent
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asked 1 hour ago
CuriousStudent
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asked 1 hour ago
CuriousStudent
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2 Answers
2
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.
answered 1 hour ago
Trevor Kafka
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
 |Â
show 1 more comment
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.
answered 45 mins ago
Rob Jeffries
65.9k7130223
add a comment |Â
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.
answered 1 hour ago
Trevor Kafka
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
 |Â
show 1 more comment
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.
answered 1 hour ago
Trevor Kafka
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
 |Â
show 1 more comment
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.
answered 1 hour ago
Trevor Kafka
490210
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.
answered 1 hour ago
Trevor Kafka
490210
answered 1 hour ago
Trevor Kafka
490210
answered 1 hour ago
Trevor Kafka
490210
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
 |Â
show 1 more comment
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
 |Â
show 1 more comment
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.
answered 45 mins ago
Rob Jeffries
65.9k7130223
add a comment |Â
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.
answered 45 mins ago
Rob Jeffries
65.9k7130223
add a comment |Â
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.
answered 45 mins ago
Rob Jeffries
65.9k7130223
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.
answered 45 mins ago
Rob Jeffries
65.9k7130223
answered 45 mins ago
Rob Jeffries
65.9k7130223
answered 45 mins ago
Rob Jeffries
65.9k7130223
answered 45 mins ago
Rob Jeffries
65.9k7130223
65.9k7130223
add a comment |Â
add a comment |Â
CuriousStudent is a new contributor. Be nice, and check out our Code of Conduct.
CuriousStudent is a new contributor. Be nice, and check out our Code of Conduct.
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