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In the definition of the electrical field we use the concept of a test charge because we state that the point charge is required for the direct application of the Coloumb's Law and its infinitesimal small magnitude ensure that it does not distort the actual field that we intend to measure. But once the electrical field is defined we directly apply F=qE even if the considered point charge is of a finite magnitude. Why don't we consider the effect of the distortion of the original field here ?

Zoltan Zimboras
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2 Answers2

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While it's good to be skeptical of approximations, it should be absolutely understandable for you to ignore this effect in the case of, say, the Earth's E/M field acting on a tiny electron beam.

You couldn't find a simple general formula for the force on a particle due to the field it creates, and so things are dealt with on a case-by-case basis.

One case is the force of attraction between a neutrally charged sphere and a point charge. It is found that the static force of attraction between the charge and the sphere is $k \frac{q^2 R a}{(a^2-R^2)^2}$, where $a$ is the distance from the center of the sphere to the point charge, $R$ is the radius of the sphere, $q$ is the charge of the point, and $k=\frac{1}{4 \pi \epsilon_0}$. (Griffiths Introduction to Electrodynamics 4ed eq 3.18 if you're interested)

The same book, problem 3.41, talks about how the same thing can happen (attraction) even for a negatively charged sphere, and uses this as evidence for $\text{C}_{60}^{--}$ (http://en.wikipedia.org/wiki/Buckminsterfullerene)

So, we do consider the effects of the point charge in some cases, but only when it's absolutely necessary. I would assume that in most cases it would absolutely break your back to calculate these effects and give little benefit in terms of accuracy. (Finding the fields/potentials from conductors with applied voltages is a hard problem and amounts to solving Laplace's equation $\nabla^2\phi=0$, possibly with complicated boundary conditions.)

In short: Solving for the full effect is necessary sometimes but most of the time it is hard and unnecessary.

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    It is worthwhile to note that if the finite test charge does not disturb the distribution of the other charges at all, then we can ignore the field created by the test charge can be ignored, since it cannot create a force on the test charge by newton's laws and the principle of superposition. – Brian Moths Jan 31 '14 at 06:40
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The reason why we consider the distortion of original field is superposition of nwe charge in system.But it is clear that in rule of superposition we should never consider the effect of any charge on itself.So after once defining the electric field we can use finite point charges to apply F=qE.