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? $\left( k = \frac{1}{4\pi \varepsilon _0} \right)$

Two charges each of charge $+10 \mu C$ are kept on $Y$-axis at $y=-a$ and $y=+a$, respectively. Another point charge $-20 \mu C$ is placed at the origin and given a small displacement $x$ $(x \ll a)$ along $X$-axis. The force acting on the point charge is ($x$ and $a$ are in metres, $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \text{ N m}^2 \text{ C}^{-2}$)

Two beads,each with charge $q$ and mass $m$,are on a horizontal,frictionless,non-conducting,circular hoop of radius $R$. One of the beads is glued to the hoop at some point,while the other one performs small oscillations about its equilibrium position along the hoop. The square of the angular frequency of the small oscillations is given by [ $\varepsilon_0$ is the permittivity of free space.]

$A$ charge $Q$ is distributed over a line of length $L$. Another point charge $q$ is placed at a distance $r$ from the centre of the line distribution. Then the force experienced by $q$ is

Two point charges $q_1 = 3 \mu \text{C}$ and $q_2 = -4 \mu \text{C}$ are placed at points $(2\hat{i} + 3\hat{j} + 3\hat{k})$ and $(\hat{i} + \hat{j} + \hat{k})$ respectively. Force on charge $q_2$ is . . . . . . $N$. (Take $\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{ SI Units}$)

Two point charges $+10 \mu C$ and $+4 \mu C$ are placed $10 \ cm$ apart in air. The work required to be done to bring them $2 \ cm$ closer is $\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \text{ SI units}\right)$. (in $J$)