In Millikan's oil drop experiment,an oil drop carrying a charge $Q$ is held stationary by a potential difference of $2400\,V$ between the plates. To keep a drop of half the radius stationary,the potential difference had to be made $600\,V$. What is the charge on the second drop?

Two charges are at a distance $d$ apart. If a copper plate (conducting medium) of thickness $d/2$ is placed between them,the effective force will be

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle $\theta$ with each other. When suspended in water,the angle remains the same. If the density of the material of the sphere is $1.5 \ g/cc$,the dielectric constant of water will be (Take density of water $= 1 \ g/cc$)

$A$ charge $Q$ is distributed over a line of length $L$. Another point charge $q$ is placed at a distance $r$ from the centre of the line distribution. Then the force experienced by $q$ is

Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges $Q$ repel each other with a force $F$ when kept apart at a distance $r$. $A$ third spherical conductor having the same radius as that of $B$ but uncharged is brought in contact with $B$,then brought in contact with $C$,and finally removed away from both. The new force of repulsion between $B$ and $C$ is

Three point charges of magnitude $5 \mu C$,$0.16 \mu C$,and $0.3 \mu C$ are located at the vertices $A$,$B$,and $C$ of a right-angled triangle whose sides are $AB = 3 \, cm$,$BC = 3 \sqrt{2} \, cm$,and $CA = 3 \, cm$. Point $A$ is the right-angle corner. Calculate the magnitude of the net electrostatic force (in $N$) experienced by the charge at point $A$ due to the other two charges.