In bringing an electron towards another electron,the electrostatic potential energy of the system

Two point charges $q_1 = 6 \mu\text{C}$ and $q_2 = 4 \mu\text{C}$ are kept at points $A$ and $B$ in air where distance $AB = 10 \text{cm}$. What is the increase in potential energy of the system when $q_2$ is moved towards $q_1$ by $2 \text{cm}$? $(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{SI units})$

In the figure,the value of $Q$ so that the electrostatic potential energy of the system becomes zero is

Charges $Q$,$+q$,and $+q$ are placed at the vertices of an isosceles right-angled triangle as shown in the figure. If the total electrostatic potential energy of the system is zero,then $Q = $ . . . . . . .

When a negative charge is taken at a height from the Earth's surface,its potential energy:

Three charges $-q, Q$ and $-q$ are placed at equal distances on a straight line. If the total potential energy of the system of three charges is zero,then the ratio $\frac{Q}{q}$ is