Potential by Assembling Charges The 2019 Stack Overflow Developer Survey Results Are...
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Potential by Assembling Charges
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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?
$begingroup$
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
asked 1 hour ago
Kushal T.Kushal T.
366
$endgroup$
add a comment |
$begingroup$
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
asked 1 hour ago
Kushal T.Kushal T.
366
$endgroup$
add a comment |
$begingroup$
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
asked 1 hour ago
Kushal T.Kushal T.
366
$endgroup$
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
asked 1 hour ago
Kushal T.Kushal T.
366
asked 1 hour ago
Kushal T.Kushal T.
366
edited 16 mins ago
Kushal T.
asked 1 hour ago
Kushal T.Kushal T.
366
asked 1 hour ago
Kushal T.Kushal T.
366
asked 1 hour ago
Kushal T.Kushal T.
366
366
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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(
answered 1 hour ago
TojrahTojrah
2077
$endgroup$
$begingroup$
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.
$endgroup$
– Kushal T.
32 mins ago
add a comment |
$begingroup$
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
answered 59 mins ago
Nobody recognizeableNobody recognizeable
647617
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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(
answered 1 hour ago
TojrahTojrah
2077
$endgroup$
$begingroup$
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.
$endgroup$
– Kushal T.
32 mins ago
add a comment |
$begingroup$
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(
answered 1 hour ago
TojrahTojrah
2077
$endgroup$
$begingroup$
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.
$endgroup$
– Kushal T.
32 mins ago
add a comment |
$begingroup$
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(
answered 1 hour ago
TojrahTojrah
2077
$endgroup$
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(
answered 1 hour ago
TojrahTojrah
2077
answered 1 hour ago
TojrahTojrah
2077
answered 1 hour ago
TojrahTojrah
2077
answered 1 hour 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.
$endgroup$
– Kushal T.
32 mins ago
add a comment |
$begingroup$
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.
$endgroup$
– Kushal T.
32 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.
$endgroup$
– Kushal T.
32 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.
$endgroup$
– Kushal T.
32 mins ago
add a comment |
$begingroup$
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
answered 59 mins ago
Nobody recognizeableNobody recognizeable
647617
$endgroup$
add a comment |
$begingroup$
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
answered 59 mins ago
Nobody recognizeableNobody recognizeable
647617
$endgroup$
add a comment |
$begingroup$
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
answered 59 mins ago
Nobody recognizeableNobody recognizeable
647617
$endgroup$
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
answered 59 mins ago
Nobody recognizeableNobody recognizeable
647617
edited 38 mins ago
answered 59 mins ago
Nobody recognizeableNobody recognizeable
647617
answered 59 mins ago
Nobody recognizeableNobody recognizeable
647617
answered 59 mins ago
Nobody recognizeableNobody recognizeable
647617
647617
add a comment |
add a comment |
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