A charge $q$ is placed at the centre of the line joining two equal charges $Q$. The system of the three charges will be in equilibrium, if $q$ is equal to
$ - \frac{Q}{2}$
$ - \frac{Q}{4}$
$ + \frac{Q}{4}$
$ + \frac{Q}{2}$
The diagrams depict four different charge distributions. All the charged particles are at same distance from origin $(i.e. OA = OB = OC = OD)$ $F_1$ , $F_2$ , $F_3$ and $F_4$ are the magnitude of electrostatic force experienced by a point charge $q_0$ kept at origin in figure $-1$ , figure $-2$ , figure $-3$ and figure $-4$ respectively. Choose the correct statement.
The ratio of electrostatic and gravitational forces acting between electron and 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 = $1.6 \times {10^{ - 27}}\,kg,$ $\,G = 6.7 \times {10^{ - 11}}\,N{m^2}/k{g^2})$
Two charges $\mathrm{q}$ and $-3\mathrm{q}$ are placed fixed on $x-$ axis separated by distance $\mathrm{'d'}$. Where should a third charge $2\mathrm{q}$ be placed such that it will not experience any force ?
Three charges are placed as shown in figure. The magnitude of $q_1$ is $2.00\, \mu C$, but its sign and the value of the charge $q_2$ are not known. Charge $q_3$ is $+4.00\, \mu C$, and the net force on $q_3$ is entirely in the negative $x-$ direction. As per the condition given the sign of $q_1$ and $q_2$ will be
A total charge $Q$ is broken in two parts ${Q_1}$ and ${Q_2}$ and they are placed at a distance $R$ from each other. The maximum force of repulsion between them will occur, when