Electric Charge...

In Coulumb's law, the constant of proportionality, $$ K = \dfrac{1}{ 4 \pi \epsilon_0 } $$ has units

Two charges equal in magnitude and opposite in polarity are placed at a certain distance apart and force acting between them is $$F$$. If $$75$$% charge of one is transferred to another, then the force between the charges becomes

The force between two charges 0.06 m apart is $$5\ N$$. If each charge is moved towards the other by $$0.01\ m$$, then the force between them will become

A square surface of side $$L$$ metre in the plane of the paper is placed in a uniform electric field $$E(V/m)$$ acting along the same place at an angle $$\theta$$ with the horizontal side of the square as shown in figure. The electric flux linked to the surface in unit of $$V-m$$, is

Two point charges of $$20 \mu C$$ and $$80 \mu C$$ are $$10cm$$ apart. Where will the electric field intensity be zero on the line joining the charges from $$20\mu C$$ charge? (in m)

Two fixed charges A and B of $$5\mu C$$ is separated by a distance of 6 metre. C is the midpoint of the line joining A and B. A charge 'Q' of $$-5\mu C$$ is shot perpendicular to the line joining A and B through C with a kinetic energy of 0.06 J The charge Q comes at rest at a point D. The distance CD is

A spherical shell of radius $$R$$ has two holes $$A$$ and $$B$$ through which a ring having charge positive through without touching the shell. The flux through spherical shell is $$\dfrac{q}{3 \varepsilon_0}$$. Find the radius of the ring:-

Equal charges of each $$2\mu C$$ are placed at a point $$x=0,2,4,$$ and $$8$$ $$cm$$ on the $$x-$$axis. The force experienced by the charge at $$x=2\ cm$$ is equal 

N fundamental charges each of the charge '$$q$$' are to be distributed as two point charges separated by a fixed distance, then the maximum to minimum force bears a ratio ( N is even and greater than $$2$$)