Electric Charge...

Two charges are placed in vacuum at a distance 'd' apart. The force between them is 'F'. If a medium of a dielectric constant 4 is introduced between them, the force will be:

In Coulomb's law, the constant of proportionality K has the units

If two charges of 1 coulomb are placed 1 km apart, the force between them will be:

Two positively charged particles each of mass $$1.7 \times 10^{-27}$$ kg and carrying a charge of $$1.6 \times 10^{-19}$$ C are placed at a distance $$r$$ apart. If each one experiences a repulsive force equal to its weight, then $$r$$ is :

Cathode rays gain K.E. when they are accelerated by an electric field. If cathode rays are subjected to the action of magnetic field, then their

A point charge $$+10\mu C$$ is at a distance $$5\ cm$$ directly above the centre of a square of side $$10\ cm$$, as shown in the figure. What is the magnitude of the electric flux through the square?

Find the minimum force between the two electrons around He nucleus. Assume radius of He nucleus $$=$$ 6.8 A$$^o$$.

A charged particle moves with a speed v in a circular path of radius r around a long uniformly charged conductor then :

Two particles having charges $${ q }_{ 1 }$$ and $${ q }_{ 2 }$$ exert a force $$F$$ on each other, when they are placed at a certain distance. What will be the force between them, if the distance between them is reduced to half and the charge on each particle is doubled ?

Electric flux through a surface of area $$100 m^{ 2 }$$ lying in the xy plane is (in Vm) if $$\overrightarrow { E } =\hat { i } +\sqrt { 2 } \hat { j } +\sqrt { 3 } \hat { k } $$