Three charges each of magnitude $q$ are placed at the corners of an equilateral triangle. The electrostatic force on the charge $Q$ placed at the center is (each side of the triangle is $L$):

Three charges,each $+Q$,are placed at the vertices of an equilateral triangle of side $a$. What is the net electrostatic force on any one charge? $\left( k = \frac{1}{4\pi \varepsilon _0} \right)$

$A$ 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)

$A$ 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:

An electron is moving around the nucleus of a hydrogen atom in a circular orbit of radius $r$. The Coulomb force $\overrightarrow{F}$ between the two is (where $K = \frac{1}{4\pi\varepsilon_0}$):

Three point charges of magnitude $5 \mu C$,$0.16 \mu C$,and $0.3 \mu C$ are located at the vertices $A$,$B$,and $C$ of a right-angled triangle whose sides are $AB = 3 \, cm$,$BC = 3 \sqrt{2} \, cm$,and $CA = 3 \, cm$. Point $A$ is the right-angle corner. Calculate the magnitude of the net electrostatic force (in $N$) experienced by the charge at point $A$ due to the other two charges.