Two free point charges $+q$ and $+4q$ are a distance $R$ apart. $A$ third charge is so placed that the entire system is in equilibrium. Then the third charge is :-

Two identical charged particles,each having a mass $10 \, g$ and charge $2.0 \times 10^{-7} \, C$,are placed on a horizontal table with a separation of $L$ between them such that they stay in limiting equilibrium. If the coefficient of friction between each particle and the table is $0.25$,find the value of $L$ in $cm$. [Use $g = 10 \, m/s^2$]

$A$ charge $q$ is placed in the middle of a line joining two equal and like point charges $Q$. For what value of $q$ will the system be in equilibrium?

Two identical charged spheres suspended from a common point by two massless strings of length $l$ are initially a distance $d$ $(d \ll l)$ apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result,the spheres approach each other with a velocity $v$. Then,as a function of distance $x$ between them,

There is a uniform spherically symmetric surface charge density at a distance $R_0$ from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed $V(R(t))$ of the distribution as a function of its instantaneous radius $R(t)$ is

Six point charges are placed at the vertices of a regular hexagon as shown in the figure. Three of the charges are $+q$ and the other three are $-q$. Starting from $P$ and moving clockwise,the electric field at the center $O$ is twice the electric field due to a single charge $+q$ at $R$. Which of the following arrangements of charges at $P, Q, R, S, T, U$ is correct?