When the distance between the charged particles is halved,the force between them becomes

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 point charges of $+2 \, \mu \text{C}$ and $+6 \, \mu \text{C}$ repel each other with a force of $12 \, \text{N}$. If a charge of $-4 \, \mu \text{C}$ is added to each,the force will be:

As shown in the figure,a configuration of two equal point charges $(q_0 = +2 \mu C)$ is placed on an inclined plane. The mass of each point charge is $20 \, g$. Assume that there is no friction between the charge and the plane. For the system of two point charges to be in equilibrium (at rest),the height $h = x \times 10^{-3} \, m$. The value of $x$ is $..........$. (Take $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, Nm^2 C^{-2}, g = 10 \, ms^{-2}$)

Three charges $+Q, q, +Q$ are placed respectively at distances $0, \frac{d}{2},$ and $d$ from the origin on the $x$-axis. If the net force experienced by the charge $+Q$ placed at $x = 0$ is zero,then the value of $q$ is:

An oil drop of $12$ excess electrons is held stationary under a constant electric field of $2.55 \times 10^{4} \; N \, C^{-1}$ (Millikan's oil drop experiment). The density of the oil is $1.26 \; g \, cm^{-3}$. Estimate the radius of the drop. $(g = 9.81 \; m \, s^{-2}; e = 1.60 \times 10^{-19} \; C)$