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

• Coulomb's law is the law of electrostatics force between electric charges.
  Statement
  

  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.
  
  
  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 r^ˆ₂₁ and r^ˆ ₁₂ can be defined as
r^ˆ₂₁ = r₂₁/r directed from q₁ to q₂
r^ˆ₁₂ = 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
  (ε₀ = 9 × 10 ^-12 C²/Nm²)
  
  

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