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Coulomb's Law


The force $\vec{F}$ between two charges $q_1$ and $q_2$ placed in free space and separated by a distance $r$ is given by \begin{align} \vec{F}=\frac{1}{4\pi\epsilon_0}\frac{q_1 q_2}{r^2}\hat{r} \end{align} where $\epsilon_0$ is the permittivity of free space. The force is repulsive if charges are of the same type and it is attractive if charges are of different type.


Problems from IIT JEE

Problem (IIT JEE 1987): A charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium if $q$ is equal to,

  1. $-Q/2$
  2. $-Q/4$
  3. $+Q/4$
  4. $+Q/2$

Solution: Let the separation between two particles of charges $Q$ be $2a$.


Coulomb's forces on the charge $q$ due to other two charges are equal and opposite. Hence, charge $q$ is always at equilibrium irrespective of its sign and magnitude. Coulomb's force on charge $Q$ due to another charge $Q$ is repulsive in nature and has magnitude $F_Q=Q^2/(16\pi\epsilon_0 a^2)$. For the charge $Q$ to be in equilibrium, Coulomb's force on it due to charge $q$ should be attractive and of magnitude $F_Q$ i.e., \begin{align} Q^2/(16\pi\epsilon_0 a^2)=-{Qq}/{(4\pi\epsilon_0 a^2)}, \end{align} which gives $q=-Q/4$.

Limitations on Coulomb's law


  1. Electric field
JEE Physics Solved Problems in Mechanics