A small particle carrying a negative charge of $1.6 \times 10^{-19} \mathrm{C}$ is suspended in equilibrium between two horizontal metal plates 8 cm apart having a potential difference of 980 V across them. The mass of the particle is $\left[\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2\right]$

Charges $3 \mathrm{Q}, \mathrm{q}$ and Q are placed along x -axis at positions $\mathrm{x}=0, \mathrm{x}=\frac{1}{3}$ and $\mathrm{x}=1$ respectively. When the force on charge Q is zero, the value of $q$ is

If a unit charge is taken from one point to another point over an equipotential surface, then

Two point charges $\mathrm{q}_1=6 \mu \mathrm{C}$ and $\mathrm{q}_2=4 \mu \mathrm{C}$ are kept at points $A$ and $B$ in air where $A B=10 \mathrm{~cm}$. What is the increase in potential energy of the system when $\mathrm{q}_2$ is moved towards $\mathrm{q}_1$ by 2 cm ?

$$\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \text { SI units }\right)$$