Two electrons are separated by a distance of $1\,\mathring{A}$. What is the Coulomb force between them?

Equal charges $q$ are placed at the four corners $A, B, C, D$ of a square of side length $a$. The magnitude of the net electrostatic force on the charge at $B$ is:

When two identical point charges are placed at a distance of $5 \, cm$, they experience a repulsive force of $0.144 \, N$. The value of each charge in microcoulombs $(\mu C)$ is:

$5$ 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 positive ions,each carrying a charge $q,$ are separated by a distance $d.$ If $F$ is the force of repulsion between the ions,the number of electrons missing from each ion will be ($e$ being the charge on an electron).

$A$ point charge $q_1$ exerts a force $F$ on a point charge $q_2$. If a third charge $q_3$ is brought near the charge $q_2$,then the force exerted by $q_1$ on $q_2$ will be: