electrostatics force : Frictional electricity : Coulomb's Law

3. Frictional Electricity

  • If we pass a comb through hairs, comb becomes electrically charged and can attract small pieces of paper.
  • Many such solid materials are known which on rubbing attract light objects like light feather, bits of papers, straw etc.
  • Explanation of appearance of electric charge on rubbing is simple.
  • Material bodies consists of large number of electrons and protons in equal number and hence is in neutral in their normal state. But when the body is rubbed for example when a glass rod is rubbed with silk cloth, electrons are transferred from glass rod to silk cloth. The glass rod becomes positively charged and the silk cloth becomes negatively charged as it receives extra electrons from the glass rod.
  • In this case rod after rubbing, comb after passing through dry hairs becomes electrified and these are the example of frictional electricity.

4. Coulumb's law for electro force

What is electrostatics force ?

Your browser does not support SVG

  • Coulomb's law is the law of electrostatics force between electric charges.
    It states that two stationary point charges q1 and q2 repel or attract each other with a electrostatics force F which is directly proportional to the product of charges and inversely proportional to the square of distance between them.
    This dependence can be expressed by writing $$F\alpha \frac{q_{1}q_{2}}{r^{2}}\tag{1}$$
  • These electrostatics forces are attractive for unlike charges and repulsive for like charges .
  • We now try to express Coulomb's law in vector form for more clarity of magnitude and direction of electrostatics forces.
  • Consider two point charges q1 and q2 at points with position vector r1 and r2 with respect to the origin.
    Position vector of one charge with respect to another charge

    vector r21= r2 - r1 is the difference between r2 and r1 and the distance of separation r is the magnitude of vector r21.
    point wise it can be written as
    r1 = position vector of charge q1 with respect to origin
    r2 = position vector of charge q2 with respect to origin
    r21 = vector from 1 to 2 (r2 - r1)
    r12 = -r21 = vector from 2 to 1 (r1 - r2)
    r = r12 = r21 = distance between 1 and 2.

    Coulomb's law for electrostatics force can then be expressed as
    $$\boldsymbol{F_{21}}=\frac{kq_{1}q_{2}\boldsymbol{r_{21}}}{r{3}}\tag{2a}$$ and, F12 = force on q1 due to q2
    $$ \boldsymbol{F_{12}}=\frac{kq_{1}q_{2}\boldsymbol{r_{12}}}{r_{3}}=-\boldsymbol{F_{21}}\tag{2b}$$ where, k is some proportionality constant and from these two equation we see that electric forces exerted by two charges on each other are equal in magnitude but are opposite in direction.
Special Case
    for simplicity we can choose q1 being placed at origin
    r1 = 0
    and if we write r2 = r the position vector of q2 then
    F21 = force on q2 due to q1
    $$\boldsymbol{F_{21}}=\frac{kq_{1}q_{2}\boldsymbol{r}}{r_{3}}\tag{3a}$$ F12 = force on q1 due to q2
    $$\boldsymbol{F_{12}}=\frac{-kq_{1}q_{2}\boldsymbol{r}}{r_{3}}\tag{3b}$$ unit vector rˆ21 and rˆ 12 can be defined as
    rˆ21 = r21/r directed from q1 to q2
    rˆ12 = r12/r directed from q2 to q1 (4)
           = -r21/r
    force can now be written in terms of unit vector given as follows
    $$\boldsymbol{F_{21}}=\frac{kq_{1}q_{2}\boldsymbol{\widehat{r}_{21}}}{r^{2}}\tag{5a}$$ $$\boldsymbol{F_{12}}=\frac{kq_{1}q_{2}\boldsymbol{\widehat{r}_{12}}}{r^{2}}\tag{5b}$$ from this we can immediately find factors giving magnitude and the directions

  • in equation (2) we find a positive constant K and experimentally found value of k is
    K = 8.98755 × 10 9 Nm2/C2
    K ≅ 9 × 10 9 Nm2/C2
    sometimes K is written as 1/4π ε0 where ε0 is the permittivity of the vacuum whose value is
    K = 1/4πε0
    0 = 9 × 10 -12 C2/Nm2)

Class 12 Maths Class 12 Physics