Is it possible that an electric field exists like $E=kx$?
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Is it possible for an electric field $vec{E}$ to exist such that its electric field intensity increases continuously, something like $E=kx$, while all the $vec{E}$ pointing in the same direction?
electrostatics electric-fields charge
edited Mar 11 at 15:36
exp ikx
639220
asked Mar 11 at 11:34
The Angel of ElistonThe Angel of Eliston
304
$endgroup$
add a comment |
$begingroup$
Is it possible for an electric field $vec{E}$ to exist such that its electric field intensity increases continuously, something like $E=kx$, while all the $vec{E}$ pointing in the same direction?
electrostatics electric-fields charge
edited Mar 11 at 15:36
exp ikx
639220
asked Mar 11 at 11:34
The Angel of ElistonThe Angel of Eliston
304
$endgroup$
add a comment |
$begingroup$
Is it possible for an electric field $vec{E}$ to exist such that its electric field intensity increases continuously, something like $E=kx$, while all the $vec{E}$ pointing in the same direction?
electrostatics electric-fields charge
edited Mar 11 at 15:36
exp ikx
639220
asked Mar 11 at 11:34
The Angel of ElistonThe Angel of Eliston
304
$endgroup$
Is it possible for an electric field $vec{E}$ to exist such that its electric field intensity increases continuously, something like $E=kx$, while all the $vec{E}$ pointing in the same direction?
electrostatics electric-fields charge
electrostatics electric-fields charge
edited Mar 11 at 15:36
exp ikx
639220
asked Mar 11 at 11:34
The Angel of ElistonThe Angel of Eliston
304
edited Mar 11 at 15:36
exp ikx
639220
asked Mar 11 at 11:34
The Angel of ElistonThe Angel of Eliston
304
edited Mar 11 at 15:36
exp ikx
639220
edited Mar 11 at 15:36
exp ikx
639220
edited Mar 11 at 15:36
exp ikx
639220
639220
asked Mar 11 at 11:34
The Angel of ElistonThe Angel of Eliston
304
asked Mar 11 at 11:34
The Angel of ElistonThe Angel of Eliston
304
asked Mar 11 at 11:34
The Angel of ElistonThe Angel of Eliston
304
304
add a comment |
add a comment |
1 Answer
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Yes, that is possible, but only in a space filled with charge (of such
a concentration as to produce the effect). Such a field has divergence
(which means it cannot happen in empty space) but no curl (which would
necessitate currents as well as charges).
answered Mar 11 at 11:44
Whit3rdWhit3rd
7,20321629
$endgroup$
$begingroup$
Interestingly the divergence of this field is $k$ which is the same as the divergence of a field $E = ky$. Meaning that they both give the same charge distributions. Presumably it is boundary conditions that establishes the difference in the two case, but just now I'm having trouble visualizing it.
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– dmckee♦
Mar 11 at 17:44
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, that is possible, but only in a space filled with charge (of such
a concentration as to produce the effect). Such a field has divergence
(which means it cannot happen in empty space) but no curl (which would
necessitate currents as well as charges).
answered Mar 11 at 11:44
Whit3rdWhit3rd
7,20321629
$endgroup$
$begingroup$
Interestingly the divergence of this field is $k$ which is the same as the divergence of a field $E = ky$. Meaning that they both give the same charge distributions. Presumably it is boundary conditions that establishes the difference in the two case, but just now I'm having trouble visualizing it.
$endgroup$
– dmckee♦
Mar 11 at 17:44
add a comment |
$begingroup$
Yes, that is possible, but only in a space filled with charge (of such
a concentration as to produce the effect). Such a field has divergence
(which means it cannot happen in empty space) but no curl (which would
necessitate currents as well as charges).
answered Mar 11 at 11:44
Whit3rdWhit3rd
7,20321629
$endgroup$
$begingroup$
Interestingly the divergence of this field is $k$ which is the same as the divergence of a field $E = ky$. Meaning that they both give the same charge distributions. Presumably it is boundary conditions that establishes the difference in the two case, but just now I'm having trouble visualizing it.
$endgroup$
– dmckee♦
Mar 11 at 17:44
add a comment |
$begingroup$
Yes, that is possible, but only in a space filled with charge (of such
a concentration as to produce the effect). Such a field has divergence
(which means it cannot happen in empty space) but no curl (which would
necessitate currents as well as charges).
answered Mar 11 at 11:44
Whit3rdWhit3rd
7,20321629
$endgroup$
Yes, that is possible, but only in a space filled with charge (of such
a concentration as to produce the effect). Such a field has divergence
(which means it cannot happen in empty space) but no curl (which would
necessitate currents as well as charges).
answered Mar 11 at 11:44
Whit3rdWhit3rd
7,20321629
answered Mar 11 at 11:44
Whit3rdWhit3rd
7,20321629
answered Mar 11 at 11:44
Whit3rdWhit3rd
7,20321629
answered Mar 11 at 11:44
Whit3rdWhit3rd
7,20321629
7,20321629
$begingroup$
Interestingly the divergence of this field is $k$ which is the same as the divergence of a field $E = ky$. Meaning that they both give the same charge distributions. Presumably it is boundary conditions that establishes the difference in the two case, but just now I'm having trouble visualizing it.
$endgroup$
– dmckee♦
Mar 11 at 17:44
add a comment |
$begingroup$
Interestingly the divergence of this field is $k$ which is the same as the divergence of a field $E = ky$. Meaning that they both give the same charge distributions. Presumably it is boundary conditions that establishes the difference in the two case, but just now I'm having trouble visualizing it.
$endgroup$
– dmckee♦
Mar 11 at 17:44
$begingroup$
Interestingly the divergence of this field is $k$ which is the same as the divergence of a field $E = ky$. Meaning that they both give the same charge distributions. Presumably it is boundary conditions that establishes the difference in the two case, but just now I'm having trouble visualizing it.
$endgroup$
– dmckee♦
Mar 11 at 17:44
$begingroup$
Interestingly the divergence of this field is $k$ which is the same as the divergence of a field $E = ky$. Meaning that they both give the same charge distributions. Presumably it is boundary conditions that establishes the difference in the two case, but just now I'm having trouble visualizing it.
$endgroup$
– dmckee♦
Mar 11 at 17:44
add a comment |
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