The ratio of gravitational force and electrostatic repulsive force between two electrons is approximately (gravitational constant $= 6.7 \times 10^{-11} \, Nm^2/kg^2$,mass of an electron $= 9.1 \times 10^{-31} \, kg$,charge on an electron $= 1.6 \times 10^{-19} \, C$).

$5$ 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. The charge at $P_1$ is removed. The acceleration of the central charge is [Given,$O P_1=O P_2=O P_3=O P_4=O P_5=1 \,m, \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \,N \cdot m^2/C^2$]

Two identical particles of mass $m$ and charge $q$ are shot at each other from a very great distance with an initial speed $v$. The distance of closest approach of these charges is

$A$ charge $+Q$ is placed at each of the diagonally opposite corners of a square. $A$ charge $-q$ is placed at each of the other diagonally opposite corners as shown. If the net electrical force on $+Q$ is zero,then $\frac{+Q}{-q}$ is equal to

$A$ charge $Q$ is divided into two parts $Q_1$ and $Q_2$. These charges are placed at a distance $R$. What will be the values of $Q_1$ and $Q_2$ for the maximum repulsive force between them?

As shown in the figure,charges $+q$ each are placed at the four vertices of a square of side length $a$. The Coulomb force acting on the charge placed at vertex $D$ is . . . . . . .