Three charges $Q$, $+q$ and $+q$ are placed at the vertices of an equilateral triangle of side $l$ as shown in the figure. If the net electrostatic energy of the system is zero, then $Q$ is equal to:

Two charges $q_1$ and $q_2$ are placed $30~\text{cm}$ apart, as shown in the figure. A third charge $q_3$ is moved along the arc of a circle of radius $40~\text{cm}$ from $C$ to $D.$ The change in the potential energy of the system is $\dfrac{q_{3}}{4 \pi \varepsilon_{0}} k,$ where $k$ is:

If $50~\text{J}$ of work must be done to move an electric charge of $2~\text{C}$ from a point where the potential is $-10~\text {volts}$ to another point where the potential is $\text{V volts}$, then the value of $\mathrm{V}$ is:
1. $5~\text {volts}$
2. $-15~\text {volts}$
3. $+15~\text {volts}$
4. $+10~\text {volts}$

Three charges, each $+q$, are placed at the corners of an equilateral triangle $ABC$ of sides $BC$, $AC$, and $AB$. $D$ and $E$ are the mid-points of $BC$ and $CA$. The work done in taking a charge $Q$ from $D$ to $E$ is:

A bullet of mass $2~\text {gm}$ has a charge of $2~\mu\text{C}.$ Through what potential difference must it be accelerated, starting from rest, to acquire a speed of $10~\text{m/s}?$
1. $50~\text {kV}$
2. $5~\text {V}$
3. $50~\text {V}$
4. $5~\text {kV}$

Four electric charges $+ q,$ $+ q,$ $- q$ and $- q$ are placed at the corners of a square of side $2L$ (see figure). The electric potential at the point $A$, mid-way between the two charges $+ q$ and $+ q$ is:
              
1. $\frac{1}{4 \pi\varepsilon_{0}} \frac{2 q}{L} \left(1 + \frac{1}{\sqrt{5}}\right)$
2. $\frac{1}{4 \pi\varepsilon_{0}} \frac{2 q}{L} \left(1 - \frac{1}{\sqrt{5}}\right)$
3. zero
4. $\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q}{L} \left(1 + \sqrt{5}\right)$

The increasing order of the electrostatic potential energies for the given system of charges is given by: