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 \(-10\) volts to another point where the potential is \(\mathrm{V}\) volts, then the value of \(\mathrm{V}\) is:
1. \(5\) volts
2. \(-15\) volts
3. \(+15\) volts
4. \(+10\) 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 \(2\) g is having a charge of \(2\) µC. Through what potential difference must it be accelerated, starting from rest, to acquire a speed of \(10\) m/s?
1. \(50\) kV
2. \(5\) V
3. \(50\) V
4. \(5\) 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 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: