Two charges $ + 4e$ and $ + e$ are at a distance $x$ apart. At what distance, a charge $q$ must be placed from charge $ + e$ so that it is in equilibrium

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 = $1.6 \times {10^{ - 27}}\,kg,$ $\,G = 6.7 \times {10^{ - 11}}\,N{m^2}/k{g^2})$

Write some important points for vector form of Coulomb’s law.

$5$ charges each of magnitude $10^{-5} \,C$ and mass $1 \,kg$ are placed (fixed) symmetrically about a movable central charge of magnitude $5 \times 10^{-5} \,C$ and mass $0.5 \,kg$ as shown in the figure given below. The charge at $P_1$ is removed. The acceleration of the central charge is [Given, $\left.O P_1=O P_2=O P_3=O P_4=O P_5=1 m , \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9\right]$

Two similar spheres having $ + \,q$ and $ - \,q$ charge are kept at a certain distance. $F$ force acts between the two. If in the middle of two spheres, another similar sphere having $ + \,q$ charge is kept, then it experience a force in magnitude and direction as

Two equal positive point charges are separated by a distance $2 a$. The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge $q_0$ becomes maximum is $\frac{a}{\sqrt{x}}$. The value of $x$ is $................$