The acceleration of an electron due to the mutual attraction between the electron and a proton when they are $1.6 \; \mathring{A}$ apart is,$(m_{e} \simeq 9 \times 10^{-31} \; kg, e = 1.6 \times 10^{-19} \; C)$. (Take $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \; Nm^{2} C^{-2}$)

Two charges $q$ and $-3q$ are placed fixed on the $x$-axis separated by a distance $d$. Where should a third charge $2q$ be placed such that it will not experience any force?

Two charges are placed as shown in the figure. Where should a third charge be placed so that it remains at rest?

$A$ particle of mass $M$ and charge $q$ is placed at the midpoint between two fixed charges each of magnitude $Q$,separated by a distance $2d$. The system is collinear as shown in the figure. If the particle is displaced by a small distance $x$ $(x \ll d)$ along the line joining the two charges and released,it will oscillate about the mean position with a time period $T$. ($\varepsilon_{0}$ is the permittivity of free space)

Four charges equal to $-Q$ are placed at the four corners of a square and a charge $q$ is at its centre. If the system is in equilibrium,the value of $q$ is

$A$ charge $Q$ is divided into two parts $q$ and $Q - q$. If the Coulomb repulsion between them when they are separated by a distance $r$ is to be maximum,the ratio of $\frac{Q}{q}$ should be: