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Why is the electric field between two parallel plates uniform?
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According to sources online (eg HyperPhysics) the electric field strength around a point charge is $$E=kfracQr^2$$
This must means that the further you get away, the electric field should decrease with the square of the radius right?
But when these charges are placed into parallel plates somehow these will produce a uniform electric field? How I currently understand the physics can be summarized in the graph below of the electric field strength vs distance (Red and black lines are the electric field strengths of each plate and the blue is the resultant).
It seems pretty clear this wouldn't be a uniform electric field. What happens such that the electric field strength is changed into a uniform field? I initially thought integration, but wouldn't that still give an inverse relationship?
electrostatics electric-fields
asked 2 hours ago
John Hon
23129
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up vote
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down vote
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According to sources online (eg HyperPhysics) the electric field strength around a point charge is $$E=kfracQr^2$$
This must means that the further you get away, the electric field should decrease with the square of the radius right?
But when these charges are placed into parallel plates somehow these will produce a uniform electric field? How I currently understand the physics can be summarized in the graph below of the electric field strength vs distance (Red and black lines are the electric field strengths of each plate and the blue is the resultant).
It seems pretty clear this wouldn't be a uniform electric field. What happens such that the electric field strength is changed into a uniform field? I initially thought integration, but wouldn't that still give an inverse relationship?
electrostatics electric-fields
asked 2 hours ago
John Hon
23129
The intensity of light from a point on the wall that reaches your eye drops as $1/r^2$ as you slowly back away from the wall, and yet the wall doesn't appear to get darker.
– uhoh
2 mins ago
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up vote
1
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up vote
1
down vote
favorite
According to sources online (eg HyperPhysics) the electric field strength around a point charge is $$E=kfracQr^2$$
This must means that the further you get away, the electric field should decrease with the square of the radius right?
But when these charges are placed into parallel plates somehow these will produce a uniform electric field? How I currently understand the physics can be summarized in the graph below of the electric field strength vs distance (Red and black lines are the electric field strengths of each plate and the blue is the resultant).
It seems pretty clear this wouldn't be a uniform electric field. What happens such that the electric field strength is changed into a uniform field? I initially thought integration, but wouldn't that still give an inverse relationship?
electrostatics electric-fields
asked 2 hours ago
John Hon
23129
According to sources online (eg HyperPhysics) the electric field strength around a point charge is $$E=kfracQr^2$$
This must means that the further you get away, the electric field should decrease with the square of the radius right?
But when these charges are placed into parallel plates somehow these will produce a uniform electric field? How I currently understand the physics can be summarized in the graph below of the electric field strength vs distance (Red and black lines are the electric field strengths of each plate and the blue is the resultant).
It seems pretty clear this wouldn't be a uniform electric field. What happens such that the electric field strength is changed into a uniform field? I initially thought integration, but wouldn't that still give an inverse relationship?
electrostatics electric-fields
electrostatics electric-fields
asked 2 hours ago
John Hon
23129
asked 2 hours ago
John Hon
23129
asked 2 hours ago
John Hon
23129
asked 2 hours ago
John Hon
23129
asked 2 hours ago
John Hon
23129
23129
The intensity of light from a point on the wall that reaches your eye drops as $1/r^2$ as you slowly back away from the wall, and yet the wall doesn't appear to get darker.
– uhoh
2 mins ago
add a comment |Â
The intensity of light from a point on the wall that reaches your eye drops as $1/r^2$ as you slowly back away from the wall, and yet the wall doesn't appear to get darker.
– uhoh
2 mins ago
The intensity of light from a point on the wall that reaches your eye drops as $1/r^2$ as you slowly back away from the wall, and yet the wall doesn't appear to get darker.
– uhoh
2 mins ago
The intensity of light from a point on the wall that reaches your eye drops as $1/r^2$ as you slowly back away from the wall, and yet the wall doesn't appear to get darker.
– uhoh
2 mins ago
add a comment |Â
4 Answers
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The intuitive answer is the following: When you have only one infinite plate the case is the same. If the plate is infinite in lenght, then "there is no spatial scale" in this problem, there is no center and there is nothing (no physical features) that can tell you that you are closer or farther from the plate, any height would be the same. Of course you can measure the distance from the plate with a meter, but the point is that there is no features on the plate that will make one distance "different" that another. Now if you have two plates of oppossite charges it is the same, the field will be constant inside the plates and zero outside (as it cancels). This stops being true if the plates are finite, because now you have a scale: the size of the plate.
answered 1 hour ago
Wolphram jonny
9,88322451
What do you mean "there is no spatial scale"? The distance from the plate a gives spatial scale. In your answer, you merely state that you can ignore it with no justification. Why doesn't the same argument work with a point charge, or with a charge evenly distributed along an infinite wire?
– Peter Shor
12 mins ago
@PeterShor for $1/r^2$ things all that matters is solid angle, and when your plate is infinite, there is no solid-angle-scale: the plate looks the same from 1 (classical) micron away as it does one light year. Now the wire, it get's skinnier the further you move away from it--even though it's already 0 width, it goes to $0timesfracr'r$.
– JEB
3 mins ago
@JEB: thank you for supplying the justification that was completely missing in the above answer.
– Peter Shor
2 mins ago
add a comment |Â
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1
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You have to remember that the electric field is a vector and not a scalar.
Let's consider a uniform electric charge on a infinite plane, and a point charge at height $h$ from it. Let's call the point directly below the point charge $P$.
The point charge gets attracted to every point on the plane, and the attractive force to a point on the plane at distance $r$ from $P$ is proportional to
$$
frac1r^2+h^2
$$.
Now, the fraction of this force in the direction of the plane is proportional to a
$$
frachsqrtr^2+h^2
$$
fraction of the total force.
Thus, the force on the point charge in the direction of the plane is
$$
int_0^infty frach(r^2+h^2)^3/2, 2 pi r , dr
$$
This integral doesn't depend on $h$. To see this, we can make the substitution $r = h s$. We get
$$
int_0^infty frach(h^2 s^2+h^2)^3/2 ,2 pi h^2 s , ds =
int_0^infty frac1(s^2+1)^3/2 ,2 pi s , ds , ,
$$
which is $2 pi$.
So the attractive force on the point charge doesn't depend on its distance from the plane. Thus, the electric field must be constant.
answered 17 mins ago
Peter Shor
8,3352152
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up vote
0
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If you place a uniform charge density, so not a single highly non-uniform point charge, on each of the plates, then far enough away from the edges of the plates the E field will be constant. Note that the plates should be much larger than their separation.
answered 1 hour ago
my2cts
3,5922416
Could you please explain WHY this happens though? Why will it become constant when you have two parallel plates?
– John Hon
1 hour ago
One answer would be : because of Gauss's law. Another way to look at this is to consider a point P at height z above an infinite plain. $vec E$ is the vector sum of all contributions from the infinite plain. When you change z then all charges are further away, which tends to decrease the field . However, we need the vectorial sum here. The angles under which the charges are seen from P are decreasing giving an increase of the field at P. Both effects exactly cancel.
– my2cts
1 hour ago
add a comment |Â
up vote
0
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You're right !
It seems pretty clear this wouldn't be a uniform electric field.
This uniform field is only valid under certain approximations.
First the separation between plates must be very small compared to the size of the plates. This means that when we expand the exact solution in terms of powers $frac x L$ where $x$ is the distance from plates and $L$ is a measure of the size of the plates, we can neglect anything but the constant term.
Typically we also assume an infinitely large plate, so that there are no boundary effects.
In a real situation, the edges ("boundary") clearly won't have the same field as the middle of the plate. So the field cannot be uniform for finite plates. Hence we approximate with infinite plates, which works well for the field in around the middle, and in the case of small separations this works well enough for most of the plate.
When you use separations comparable in magnitude to the size of the plates, you will not get a uniform field. It's easy to see that in that case the closer you are to one plate or another, the more influence it has on the field around you. In fact when the separation becomes much larger than the size f the plates, you can approximate the plates with points and you end up with an approximation of the field between two point charges, which is definitely not uniform.
answered 10 mins ago
StephenG
5,13821324
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The intuitive answer is the following: When you have only one infinite plate the case is the same. If the plate is infinite in lenght, then "there is no spatial scale" in this problem, there is no center and there is nothing (no physical features) that can tell you that you are closer or farther from the plate, any height would be the same. Of course you can measure the distance from the plate with a meter, but the point is that there is no features on the plate that will make one distance "different" that another. Now if you have two plates of oppossite charges it is the same, the field will be constant inside the plates and zero outside (as it cancels). This stops being true if the plates are finite, because now you have a scale: the size of the plate.
answered 1 hour ago
Wolphram jonny
9,88322451
What do you mean "there is no spatial scale"? The distance from the plate a gives spatial scale. In your answer, you merely state that you can ignore it with no justification. Why doesn't the same argument work with a point charge, or with a charge evenly distributed along an infinite wire?
– Peter Shor
12 mins ago
@PeterShor for $1/r^2$ things all that matters is solid angle, and when your plate is infinite, there is no solid-angle-scale: the plate looks the same from 1 (classical) micron away as it does one light year. Now the wire, it get's skinnier the further you move away from it--even though it's already 0 width, it goes to $0timesfracr'r$.
– JEB
3 mins ago
@JEB: thank you for supplying the justification that was completely missing in the above answer.
– Peter Shor
2 mins ago
add a comment |Â
up vote
1
down vote
The intuitive answer is the following: When you have only one infinite plate the case is the same. If the plate is infinite in lenght, then "there is no spatial scale" in this problem, there is no center and there is nothing (no physical features) that can tell you that you are closer or farther from the plate, any height would be the same. Of course you can measure the distance from the plate with a meter, but the point is that there is no features on the plate that will make one distance "different" that another. Now if you have two plates of oppossite charges it is the same, the field will be constant inside the plates and zero outside (as it cancels). This stops being true if the plates are finite, because now you have a scale: the size of the plate.
answered 1 hour ago
Wolphram jonny
9,88322451
What do you mean "there is no spatial scale"? The distance from the plate a gives spatial scale. In your answer, you merely state that you can ignore it with no justification. Why doesn't the same argument work with a point charge, or with a charge evenly distributed along an infinite wire?
– Peter Shor
12 mins ago
@PeterShor for $1/r^2$ things all that matters is solid angle, and when your plate is infinite, there is no solid-angle-scale: the plate looks the same from 1 (classical) micron away as it does one light year. Now the wire, it get's skinnier the further you move away from it--even though it's already 0 width, it goes to $0timesfracr'r$.
– JEB
3 mins ago
@JEB: thank you for supplying the justification that was completely missing in the above answer.
– Peter Shor
2 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The intuitive answer is the following: When you have only one infinite plate the case is the same. If the plate is infinite in lenght, then "there is no spatial scale" in this problem, there is no center and there is nothing (no physical features) that can tell you that you are closer or farther from the plate, any height would be the same. Of course you can measure the distance from the plate with a meter, but the point is that there is no features on the plate that will make one distance "different" that another. Now if you have two plates of oppossite charges it is the same, the field will be constant inside the plates and zero outside (as it cancels). This stops being true if the plates are finite, because now you have a scale: the size of the plate.
answered 1 hour ago
Wolphram jonny
9,88322451
The intuitive answer is the following: When you have only one infinite plate the case is the same. If the plate is infinite in lenght, then "there is no spatial scale" in this problem, there is no center and there is nothing (no physical features) that can tell you that you are closer or farther from the plate, any height would be the same. Of course you can measure the distance from the plate with a meter, but the point is that there is no features on the plate that will make one distance "different" that another. Now if you have two plates of oppossite charges it is the same, the field will be constant inside the plates and zero outside (as it cancels). This stops being true if the plates are finite, because now you have a scale: the size of the plate.
answered 1 hour ago
Wolphram jonny
9,88322451
answered 1 hour ago
Wolphram jonny
9,88322451
answered 1 hour ago
Wolphram jonny
9,88322451
answered 1 hour ago
Wolphram jonny
9,88322451
9,88322451
What do you mean "there is no spatial scale"? The distance from the plate a gives spatial scale. In your answer, you merely state that you can ignore it with no justification. Why doesn't the same argument work with a point charge, or with a charge evenly distributed along an infinite wire?
– Peter Shor
12 mins ago
@PeterShor for $1/r^2$ things all that matters is solid angle, and when your plate is infinite, there is no solid-angle-scale: the plate looks the same from 1 (classical) micron away as it does one light year. Now the wire, it get's skinnier the further you move away from it--even though it's already 0 width, it goes to $0timesfracr'r$.
– JEB
3 mins ago
@JEB: thank you for supplying the justification that was completely missing in the above answer.
– Peter Shor
2 mins ago
add a comment |Â
What do you mean "there is no spatial scale"? The distance from the plate a gives spatial scale. In your answer, you merely state that you can ignore it with no justification. Why doesn't the same argument work with a point charge, or with a charge evenly distributed along an infinite wire?
– Peter Shor
12 mins ago
@PeterShor for $1/r^2$ things all that matters is solid angle, and when your plate is infinite, there is no solid-angle-scale: the plate looks the same from 1 (classical) micron away as it does one light year. Now the wire, it get's skinnier the further you move away from it--even though it's already 0 width, it goes to $0timesfracr'r$.
– JEB
3 mins ago
@JEB: thank you for supplying the justification that was completely missing in the above answer.
– Peter Shor
2 mins ago
What do you mean "there is no spatial scale"? The distance from the plate a gives spatial scale. In your answer, you merely state that you can ignore it with no justification. Why doesn't the same argument work with a point charge, or with a charge evenly distributed along an infinite wire?
– Peter Shor
12 mins ago
What do you mean "there is no spatial scale"? The distance from the plate a gives spatial scale. In your answer, you merely state that you can ignore it with no justification. Why doesn't the same argument work with a point charge, or with a charge evenly distributed along an infinite wire?
– Peter Shor
12 mins ago
@PeterShor for $1/r^2$ things all that matters is solid angle, and when your plate is infinite, there is no solid-angle-scale: the plate looks the same from 1 (classical) micron away as it does one light year. Now the wire, it get's skinnier the further you move away from it--even though it's already 0 width, it goes to $0timesfracr'r$.
– JEB
3 mins ago
@PeterShor for $1/r^2$ things all that matters is solid angle, and when your plate is infinite, there is no solid-angle-scale: the plate looks the same from 1 (classical) micron away as it does one light year. Now the wire, it get's skinnier the further you move away from it--even though it's already 0 width, it goes to $0timesfracr'r$.
– JEB
3 mins ago
@JEB: thank you for supplying the justification that was completely missing in the above answer.
– Peter Shor
2 mins ago
@JEB: thank you for supplying the justification that was completely missing in the above answer.
– Peter Shor
2 mins ago
add a comment |Â
up vote
1
down vote
You have to remember that the electric field is a vector and not a scalar.
Let's consider a uniform electric charge on a infinite plane, and a point charge at height $h$ from it. Let's call the point directly below the point charge $P$.
The point charge gets attracted to every point on the plane, and the attractive force to a point on the plane at distance $r$ from $P$ is proportional to
$$
frac1r^2+h^2
$$.
Now, the fraction of this force in the direction of the plane is proportional to a
$$
frachsqrtr^2+h^2
$$
fraction of the total force.
Thus, the force on the point charge in the direction of the plane is
$$
int_0^infty frach(r^2+h^2)^3/2, 2 pi r , dr
$$
This integral doesn't depend on $h$. To see this, we can make the substitution $r = h s$. We get
$$
int_0^infty frach(h^2 s^2+h^2)^3/2 ,2 pi h^2 s , ds =
int_0^infty frac1(s^2+1)^3/2 ,2 pi s , ds , ,
$$
which is $2 pi$.
So the attractive force on the point charge doesn't depend on its distance from the plane. Thus, the electric field must be constant.
answered 17 mins ago
Peter Shor
8,3352152
add a comment |Â
up vote
1
down vote
You have to remember that the electric field is a vector and not a scalar.
Let's consider a uniform electric charge on a infinite plane, and a point charge at height $h$ from it. Let's call the point directly below the point charge $P$.
The point charge gets attracted to every point on the plane, and the attractive force to a point on the plane at distance $r$ from $P$ is proportional to
$$
frac1r^2+h^2
$$.
Now, the fraction of this force in the direction of the plane is proportional to a
$$
frachsqrtr^2+h^2
$$
fraction of the total force.
Thus, the force on the point charge in the direction of the plane is
$$
int_0^infty frach(r^2+h^2)^3/2, 2 pi r , dr
$$
This integral doesn't depend on $h$. To see this, we can make the substitution $r = h s$. We get
$$
int_0^infty frach(h^2 s^2+h^2)^3/2 ,2 pi h^2 s , ds =
int_0^infty frac1(s^2+1)^3/2 ,2 pi s , ds , ,
$$
which is $2 pi$.
So the attractive force on the point charge doesn't depend on its distance from the plane. Thus, the electric field must be constant.
answered 17 mins ago
Peter Shor
8,3352152
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You have to remember that the electric field is a vector and not a scalar.
Let's consider a uniform electric charge on a infinite plane, and a point charge at height $h$ from it. Let's call the point directly below the point charge $P$.
The point charge gets attracted to every point on the plane, and the attractive force to a point on the plane at distance $r$ from $P$ is proportional to
$$
frac1r^2+h^2
$$.
Now, the fraction of this force in the direction of the plane is proportional to a
$$
frachsqrtr^2+h^2
$$
fraction of the total force.
Thus, the force on the point charge in the direction of the plane is
$$
int_0^infty frach(r^2+h^2)^3/2, 2 pi r , dr
$$
This integral doesn't depend on $h$. To see this, we can make the substitution $r = h s$. We get
$$
int_0^infty frach(h^2 s^2+h^2)^3/2 ,2 pi h^2 s , ds =
int_0^infty frac1(s^2+1)^3/2 ,2 pi s , ds , ,
$$
which is $2 pi$.
So the attractive force on the point charge doesn't depend on its distance from the plane. Thus, the electric field must be constant.
answered 17 mins ago
Peter Shor
8,3352152
You have to remember that the electric field is a vector and not a scalar.
Let's consider a uniform electric charge on a infinite plane, and a point charge at height $h$ from it. Let's call the point directly below the point charge $P$.
The point charge gets attracted to every point on the plane, and the attractive force to a point on the plane at distance $r$ from $P$ is proportional to
$$
frac1r^2+h^2
$$.
Now, the fraction of this force in the direction of the plane is proportional to a
$$
frachsqrtr^2+h^2
$$
fraction of the total force.
Thus, the force on the point charge in the direction of the plane is
$$
int_0^infty frach(r^2+h^2)^3/2, 2 pi r , dr
$$
This integral doesn't depend on $h$. To see this, we can make the substitution $r = h s$. We get
$$
int_0^infty frach(h^2 s^2+h^2)^3/2 ,2 pi h^2 s , ds =
int_0^infty frac1(s^2+1)^3/2 ,2 pi s , ds , ,
$$
which is $2 pi$.
So the attractive force on the point charge doesn't depend on its distance from the plane. Thus, the electric field must be constant.
answered 17 mins ago
Peter Shor
8,3352152
edited 8 mins ago
answered 17 mins ago
Peter Shor
8,3352152
answered 17 mins ago
Peter Shor
8,3352152
answered 17 mins ago
Peter Shor
8,3352152
8,3352152
add a comment |Â
add a comment |Â
up vote
0
down vote
If you place a uniform charge density, so not a single highly non-uniform point charge, on each of the plates, then far enough away from the edges of the plates the E field will be constant. Note that the plates should be much larger than their separation.
answered 1 hour ago
my2cts
3,5922416
Could you please explain WHY this happens though? Why will it become constant when you have two parallel plates?
– John Hon
1 hour ago
One answer would be : because of Gauss's law. Another way to look at this is to consider a point P at height z above an infinite plain. $vec E$ is the vector sum of all contributions from the infinite plain. When you change z then all charges are further away, which tends to decrease the field . However, we need the vectorial sum here. The angles under which the charges are seen from P are decreasing giving an increase of the field at P. Both effects exactly cancel.
– my2cts
1 hour ago
add a comment |Â
up vote
0
down vote
If you place a uniform charge density, so not a single highly non-uniform point charge, on each of the plates, then far enough away from the edges of the plates the E field will be constant. Note that the plates should be much larger than their separation.
answered 1 hour ago
my2cts
3,5922416
Could you please explain WHY this happens though? Why will it become constant when you have two parallel plates?
– John Hon
1 hour ago
One answer would be : because of Gauss's law. Another way to look at this is to consider a point P at height z above an infinite plain. $vec E$ is the vector sum of all contributions from the infinite plain. When you change z then all charges are further away, which tends to decrease the field . However, we need the vectorial sum here. The angles under which the charges are seen from P are decreasing giving an increase of the field at P. Both effects exactly cancel.
– my2cts
1 hour ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If you place a uniform charge density, so not a single highly non-uniform point charge, on each of the plates, then far enough away from the edges of the plates the E field will be constant. Note that the plates should be much larger than their separation.
answered 1 hour ago
my2cts
3,5922416
If you place a uniform charge density, so not a single highly non-uniform point charge, on each of the plates, then far enough away from the edges of the plates the E field will be constant. Note that the plates should be much larger than their separation.
answered 1 hour ago
my2cts
3,5922416
answered 1 hour ago
my2cts
3,5922416
answered 1 hour ago
my2cts
3,5922416
answered 1 hour ago
my2cts
3,5922416
3,5922416
Could you please explain WHY this happens though? Why will it become constant when you have two parallel plates?
– John Hon
1 hour ago
One answer would be : because of Gauss's law. Another way to look at this is to consider a point P at height z above an infinite plain. $vec E$ is the vector sum of all contributions from the infinite plain. When you change z then all charges are further away, which tends to decrease the field . However, we need the vectorial sum here. The angles under which the charges are seen from P are decreasing giving an increase of the field at P. Both effects exactly cancel.
– my2cts
1 hour ago
add a comment |Â
Could you please explain WHY this happens though? Why will it become constant when you have two parallel plates?
– John Hon
1 hour ago
One answer would be : because of Gauss's law. Another way to look at this is to consider a point P at height z above an infinite plain. $vec E$ is the vector sum of all contributions from the infinite plain. When you change z then all charges are further away, which tends to decrease the field . However, we need the vectorial sum here. The angles under which the charges are seen from P are decreasing giving an increase of the field at P. Both effects exactly cancel.
– my2cts
1 hour ago
Could you please explain WHY this happens though? Why will it become constant when you have two parallel plates?
– John Hon
1 hour ago
Could you please explain WHY this happens though? Why will it become constant when you have two parallel plates?
– John Hon
1 hour ago
One answer would be : because of Gauss's law. Another way to look at this is to consider a point P at height z above an infinite plain. $vec E$ is the vector sum of all contributions from the infinite plain. When you change z then all charges are further away, which tends to decrease the field . However, we need the vectorial sum here. The angles under which the charges are seen from P are decreasing giving an increase of the field at P. Both effects exactly cancel.
– my2cts
1 hour ago
One answer would be : because of Gauss's law. Another way to look at this is to consider a point P at height z above an infinite plain. $vec E$ is the vector sum of all contributions from the infinite plain. When you change z then all charges are further away, which tends to decrease the field . However, we need the vectorial sum here. The angles under which the charges are seen from P are decreasing giving an increase of the field at P. Both effects exactly cancel.
– my2cts
1 hour ago
add a comment |Â
up vote
0
down vote
You're right !
It seems pretty clear this wouldn't be a uniform electric field.
This uniform field is only valid under certain approximations.
First the separation between plates must be very small compared to the size of the plates. This means that when we expand the exact solution in terms of powers $frac x L$ where $x$ is the distance from plates and $L$ is a measure of the size of the plates, we can neglect anything but the constant term.
Typically we also assume an infinitely large plate, so that there are no boundary effects.
In a real situation, the edges ("boundary") clearly won't have the same field as the middle of the plate. So the field cannot be uniform for finite plates. Hence we approximate with infinite plates, which works well for the field in around the middle, and in the case of small separations this works well enough for most of the plate.
When you use separations comparable in magnitude to the size of the plates, you will not get a uniform field. It's easy to see that in that case the closer you are to one plate or another, the more influence it has on the field around you. In fact when the separation becomes much larger than the size f the plates, you can approximate the plates with points and you end up with an approximation of the field between two point charges, which is definitely not uniform.
answered 10 mins ago
StephenG
5,13821324
add a comment |Â
up vote
0
down vote
You're right !
It seems pretty clear this wouldn't be a uniform electric field.
This uniform field is only valid under certain approximations.
First the separation between plates must be very small compared to the size of the plates. This means that when we expand the exact solution in terms of powers $frac x L$ where $x$ is the distance from plates and $L$ is a measure of the size of the plates, we can neglect anything but the constant term.
Typically we also assume an infinitely large plate, so that there are no boundary effects.
In a real situation, the edges ("boundary") clearly won't have the same field as the middle of the plate. So the field cannot be uniform for finite plates. Hence we approximate with infinite plates, which works well for the field in around the middle, and in the case of small separations this works well enough for most of the plate.
When you use separations comparable in magnitude to the size of the plates, you will not get a uniform field. It's easy to see that in that case the closer you are to one plate or another, the more influence it has on the field around you. In fact when the separation becomes much larger than the size f the plates, you can approximate the plates with points and you end up with an approximation of the field between two point charges, which is definitely not uniform.
answered 10 mins ago
StephenG
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You're right !
It seems pretty clear this wouldn't be a uniform electric field.
This uniform field is only valid under certain approximations.
First the separation between plates must be very small compared to the size of the plates. This means that when we expand the exact solution in terms of powers $frac x L$ where $x$ is the distance from plates and $L$ is a measure of the size of the plates, we can neglect anything but the constant term.
Typically we also assume an infinitely large plate, so that there are no boundary effects.
In a real situation, the edges ("boundary") clearly won't have the same field as the middle of the plate. So the field cannot be uniform for finite plates. Hence we approximate with infinite plates, which works well for the field in around the middle, and in the case of small separations this works well enough for most of the plate.
When you use separations comparable in magnitude to the size of the plates, you will not get a uniform field. It's easy to see that in that case the closer you are to one plate or another, the more influence it has on the field around you. In fact when the separation becomes much larger than the size f the plates, you can approximate the plates with points and you end up with an approximation of the field between two point charges, which is definitely not uniform.
answered 10 mins ago
StephenG
5,13821324
You're right !
It seems pretty clear this wouldn't be a uniform electric field.
This uniform field is only valid under certain approximations.
First the separation between plates must be very small compared to the size of the plates. This means that when we expand the exact solution in terms of powers $frac x L$ where $x$ is the distance from plates and $L$ is a measure of the size of the plates, we can neglect anything but the constant term.
Typically we also assume an infinitely large plate, so that there are no boundary effects.
In a real situation, the edges ("boundary") clearly won't have the same field as the middle of the plate. So the field cannot be uniform for finite plates. Hence we approximate with infinite plates, which works well for the field in around the middle, and in the case of small separations this works well enough for most of the plate.
When you use separations comparable in magnitude to the size of the plates, you will not get a uniform field. It's easy to see that in that case the closer you are to one plate or another, the more influence it has on the field around you. In fact when the separation becomes much larger than the size f the plates, you can approximate the plates with points and you end up with an approximation of the field between two point charges, which is definitely not uniform.
answered 10 mins ago
StephenG
5,13821324
answered 10 mins ago
StephenG
5,13821324
answered 10 mins ago
StephenG
5,13821324
answered 10 mins ago
StephenG
5,13821324
5,13821324
add a comment |Â
add a comment |Â
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The intensity of light from a point on the wall that reaches your eye drops as $1/r^2$ as you slowly back away from the wall, and yet the wall doesn't appear to get darker.
– uhoh
2 mins ago