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As I was studying about the electric field and its effect I came across the idea that to find the intensity of an electric field we consider a positive unit charge particle at that point.

If we have lots of charged particles in that particular electric field then the electric field will get disturbed by these charged particles and if it happen so then how we can consider the uniform electric field in the presence of these charge particles?

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  • $\begingroup$ The electric field 'acting' on one of the particles is the superposition of the fields from those other charged particles in addition to the specified background field isn't it? $\endgroup$ – Alfred Centauri Jun 1 '18 at 12:10
  • $\begingroup$ Using a test charge $q$ the electric field is $\vec E = \frac {\text{force on charge q}}{q}$ as the charge $q$ tends to zero. $\endgroup$ – Farcher Jun 1 '18 at 13:46
  • $\begingroup$ If you have "lots" of charges particles that aren't bound to a plane and uniformly distributed, there is no uniform electric field. $\endgroup$ – Bill N Jun 2 '18 at 23:48
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Don't define electric field strength, $\vec{E}$, at point P as force on a unit charge placed at P. For one thing this gives the wrong units for $\vec{E}$ (newton instead of $\text{newton coulomb}\ ^{-1}$). For another thing, a unit charge in SI is the coulomb, and a compact charge of a coulomb (if it could be prevented from flying apart) would have a huge electric field around it. This field would no doubt displace the very charges ('source' charges) whose resultant field strength at P you are supposed to be measuring! [It would probably destroy the laboratory and beyond!]

I recommend this definition…

The electric field strength, $\vec{E}$, at a point P is defined by $$\vec{E}=\frac{\vec{F}}{q}$$in which $\vec{F}$ is the force that acts on a small 'test' charge, q, placed at P.

The units are now right. The test charge can be as small a charge as you like, and doesn't have to be positive. [If the test charge is negative, you'll have a negative quantity on the bottom line of the fraction, making the direction of $\vec{E}$ in the opposite direction from that of $\vec{F}$, that is in the same direction as the force on a positive charge! Halving the charge of the test charge will halve the force it experiences, so, whatever the value of q, provided that it's not too large, the same value will be obtained for $\vec{E}$. This is just what you'd hope for, because $\vec{E}$ is supposed to be telling you something about the environment of point P, independently of your means of measurement.]

As long as the test charge doesn't displace source charges from their positions, it doesn't matter that the test charge creates its own field. The force experienced by the test charge is the resultant only of the forces due to the source charges, so $\vec{E}$ as measured using the test charge is the resultant field at P due to the source charges.

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I think you have to solve an $N$-body problem in which you know the forces and the initial conditions (poisitions and velocities of the particles), and all you have to do is to solve $N$ equations for each time step and iterate them until you reach the time you want. When you do that you can find the position of the charge carriers at any given time and determine the electric field by summing all of the electric fields caused by the charge carriers plus the external electric field.

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  • $\begingroup$ No, your first paragraph has nothing to do with the question. And the electric field at a point in space has nothing to do with which forces are stronger or weaker. And how is the "external" electric field different from the fields caused by particles? The electric field is a vector field which is the superposition of all contributions. One can analyze how much contribution each source charge provides, but the actual electric field at a point is the vector sum of fields, excluding any charge at the point. OP is not asking about whether some contributions are negligible. $\endgroup$ – Bill N Jun 2 '18 at 17:43
  • $\begingroup$ @Bill N you are right. I should delete my first paragraph. However I think my second paragraph solves the problem. It is basically saying that if you know all of the charged particles' positions you can sum the electric fields and to find all of the charged particles' solutions at any given time via solving an N body problem. $\endgroup$ – Kaan Güven Jun 2 '18 at 22:23

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