Two positive charges of $20 \, C$ and $Q \, C$ are situated at a distance of $60 \, cm$. The neutral point between them is at a distance of $20 \, cm$ from the $20 \, C$ charge. The value of charge $Q$ is: (in $, C$)

Two small spheres,each having an equal positive charge $Q$ (Coulomb),are suspended by two insulating strings of equal length $L$ (metre) from a rigid hook. The whole setup is taken into a satellite where there is no gravity. The two balls are now held by electrostatic forces in a horizontal position. The tension in each string is then:

Two equal point charges exert a force $F$ on each other when they are placed distance $d$ apart in air. When they are placed distance $D$ apart in a medium of dielectric constant $K$, they exert the same force. The distance $D$ is equal to

If two particles $A$ and $B$ of charges $1.6 \times 10^{-19} \ C$ and $3.2 \times 10^{-19} \ C$ respectively are separated by a distance of $3 \ cm$ in air,then the magnitude of electrostatic force on particle $A$ due to particle $B$ is

If the distance between two point charges $+Q$ and $-Q$ is doubled,what will be the magnitude of the electrostatic force between them?

The ratio of electrostatic and gravitational forces acting between an electron and a proton separated by a distance $5 \times 10^{-11} \, m$ will be (Charge on electron $= 1.6 \times 10^{-19} \, C$,mass of electron $= 9.1 \times 10^{-31} \, kg$,mass of proton $= 1.6 \times 10^{-27} \, kg$,$G = 6.7 \times 10^{-11} \, N m^2/kg^2$).