Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The electric field at a point is defined as q/$r^2$. How does this definition take the nature of source and test charge into consideration. If I bring any positive/negative charge around the source charge, it brings a change in the distances involved due to forces of attraction or can I use this definition for practical purposes?

share|cite|improve this question
$q/r^2$ isn't the definition of the field, it's merely the force on a unit charge at a distance $r$ from the charge $q$. – John Rennie Jun 26 '13 at 9:01
$\dfrac{q}{r^2}$ isn't the definition of field. $k\cdot\dfrac{q}{r^2}$ gives the field at a distance $'r'$ from a point charge. – dajoker Jun 26 '13 at 9:56
@John Rennie: q/$r^2$ could also be the force per q1 or q2 or any other charge. – user19987 Jun 29 '13 at 10:55
@Vijay: K is medium dependent which i didn't want to consider to know the significance of definition itself. – user19987 Jun 29 '13 at 10:57

You don't take in account the complete definition of electric field:

$\vec{E}=\lim_{{q}_{0}\to0} \frac{kq{q}_{0}}{{r}^{ 2}}\hat{r}$

The test charge must be several orders in magnitude lesser than the charge that you want to know its electrical field. For convention must be positive, but this is only to define without ambiguity the direction of the electrical field.

If you use this in an application, then you should know what is the order of magnitude of the electrical charge that produce your electrical field and then choose a suitable charge test, if it is not the case then use the less charge that you can do. The point is that always your test charge affect the field that you want to measure and you have to diminish it.

share|cite|improve this answer
@Mahi-Dened:thanks for the new input. Your last sentence "The point is that always your test charge affect the field that you want to measure and you have to diminish it.".... Did you mean that test charge should not effect or effect the System of charges? – user19987 Oct 26 '13 at 17:13
The system of charges. Suppose that you have your system in electric equilibrium. If you introduce a new charge the system will not be in equilibrium any more (unless the test point is a equilibrium point). Then depends which instrumentation you are using and his response time. In conclusion, you always will have an uncertainty in the value of the electrical field that you are measuring. – Mahl-Deneb Oct 28 '13 at 17:37

The electric field at a distance $r$ from a point charge $q$ is given by $E=\dfrac{kq}{r^2}$.

Electric field is the force per unit positive test charge.

The test charge, by convention, is positive.

The attraction or repulsion depends on the point charge's polarity (i.e +/-).

share|cite|improve this answer
Electric field is the force per unit positive test charge as you said. Suppose you have calculated electric field at r.Now i bring +100C charge at r, what is the force that acts on +100c? according to you, it must be E*100C. But placing this 100C is going to bring a change in the electric field itself. so we have to recalculate the electric field and then find force. my question is what is the use of such definiton? – user19987 Jun 29 '13 at 10:51

How you can use it practically is, for example, in an electrostatic motor as in MEMS, in air purification systems that depend on moving dust particles due to an applied (high) electric field.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.