day-to-day life involve electrical charges of the order of µC (micro coulomb) or nC (nano coulomb). (v) In SI units, Coulomb’s law in vacuum takes the form q q π e . In a medium of permittivity e , the force between two point charges is given by q q π e . Since e > e o , the force between two point charges in a medium other than vacuum is always less than that in vacuum.
We define the relative permittivity for a given medium as e e e r = . For vacuum or air, e r = and for all other media e r > . 12th - 12th - - - - - Unit Electrostatics Solution y y y x x x Case (a) Case (b) Case (c) q q q q F F F F F F _ _ q q (a) q = + μC, q = + μC, and r = 1m. Both are positive charges.
so the force will be repulsive. Force experienced by the charge q due to q is given by q q π e Here r is the unit vector from q to q . Since q is located on the right of q , we have = and = × π e so that = × × × × × N q q q q ) =− e e (or) F =− Therefore, the electrostatic force obeys Newton’s third law. (viii) The expression for Coulomb force is true only for point charges.
But the point charge is an ideal concept. However we can apply Coulomb’s law for two charged objects whose sizes are very much smaller than the distance between them. In fact, Coulomb discovered his law by considering the