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What is point charge and test charge?Why coulomb's law is for point charges only?

  1. Jun 3, 2012 #1
    While studying electricity we always take point charges to calculate force between them according to coulomb's law.And to calculate the electric field around a point charge we take a test charge which is small.

    My question is:

    1.What is the definition of point charge and test charge?
    2.Why test charge is need to be small?
    3.Why coulombs law is valid only for point charges?
     
  2. jcsd
  3. Jun 3, 2012 #2
    Honestly I think it's a stupid way to look at the whole thing, I never understood why it's taught that way.

    Coulomb's law is valid for any distribution of charge, but when you have more than a single point charge you need to add up all the contributions of the individual charges to find the total field, either through a finite summation for finite charges, or through an integrations for a continuous distribution. The form of Coulomb's law you're familiar with is only valid for a point charge located at the origin, and the field only has a radial component, but you must remember that fields are vector fields.

    The general expression for the electric field of a point charge is:

    [tex] \vec{E} = \frac{q(\vec{r}-\vec{r'})}{|\vec{r}-\vec{r'}|^{3}}[/tex]

    Where [tex]\vec{r}[/tex] is the point at which you want know the field, and [tex]\vec{r'}[/tex] is the postion of the point charge. Notice this reduces to the old form if you place the point charge at the origin.
     
  4. Jun 3, 2012 #3

    Pengwuino

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    Gold Member

    A test charge is a charge that is small enough so that the charge configuration you are looking at is not disturbed. If you were to bring a [itex]1 \mu C[/itex] charge up against some giant [itex]1 C[/itex] charge, the smaller "test" charge would not really affect the position of the charge configuration you're studying. This leaves the equations dipole mentioned fairly easy to deal with.

    If, on the other hand, you put two [itex]1 C[/itex] charges in proximity (or any charge of similar magnitude), the charge configuration will feel a strong enough force from the "test" charge that it will begin to move. If your charges move, your positions are no longer time-independent and the system becomes far more challenging to analyze.

    Dipole explained what a point charge is and I basically am explaining what it means to be a "test" charge. A test charge is a charge small enough to leave the main charge configuration undisturbed. Coulomb's law is valid in general, but the form you typically see at an introductory level has no time-dependence and if you're not dealing with the differential version of Coulomb's law, it is not usable for objects other than point charges.
     
  5. Jun 3, 2012 #4
    I have got the answer about test charge,but have not got a clear idea about point charge.I know,whenever we calculate the field of a distribution of charges at any point we do integration to add up the individual contribution at that point.We divide the surface of any charged object into small surface areas (like σ=Q/A,Q=total charge,A=surface area,or linear charge density or volume charge density),and consider them as point charges and then integrate the total contribution.My question was why we always start our calculation with point charges,why we take charge density (σ) first to calculate the field of that charge distribution (such as long charged sheet,charged ring etc)...

    I prepared my answer earlier.I was just testing whether I am right or wrong.My conception is given below.Please tell me whether I am right or wrong.


    "Consider a charged body A,and a small test charge q.The force exerted by the test charge q on the charge distribution on body A may cause this distribution to shift around.This is specially true if body A is a conductor,on which charge (electrons) is easy to move.So the electric field around A when q is present may not be the same as when q is absent.But if q is very small,the redistribution of on charged body A is also small.So to make a complete correct distribution of electric field,we take the limit of test charge q approaches zero (q→0) and as the disturbing effect of q on the charge distribution becomes negligible.In practical calculations of the electric field produced by a charge distribution,we will consider the charge distribution to be fixed.Shifting of charge distribution in small conductors (like a point object) may be considered negligible.But in conductors those are not small,this shifting is not negligible.That's why we always consider point charges."


    (I am not a native English speaker.That's why my English may be bad.Sorry.)
     
  6. Jun 3, 2012 #5
    As dipole has mentioned in his post, you use point charges which you can then integrate to find out the electric field. This is because the field contributed by charges in each specific point in space is unique.
     
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