The angle between equipotential surface and electric lines of force is:
1. zero
2. \(180^{\circ}\)
3. \(90^{\circ}\)
4. \(45^{\circ}\)

A parallel plate condenser has a uniform electric field \(E\)(V/m) in the space between the plates. If the distance between the plates is \(d\)(m) and area of each plate is \(A(\text{m}^2)\), the energy (joule) stored in the condenser is:

An electric dipole of moment \(\vec {p} \) is lying along a uniform electric field \(\vec{E}.\) The work done in rotating the dipole by \(90^{\circ}\) is:
1. \(\sqrt{2}pE\)
2. \(\dfrac{pE}{2}\)
3. \(2pE\)
4. \(pE\)

If \(V=ar\) where \(a\) is a constant and \(r\) is the distance, then the electric field at a point will be proportional to:
1. \(r\)
2. \(r^{-1}\)
3. \(r^{-2}\)
4. \(r^{0}\)

The charge on the plates of the capacitor in a steady-state will be:

1. \(3~\mu\text{C}\)
2. \(9~\mu\text{C}\)
3. \(27~\mu\text{C}\)
4. \(36~\mu\text{C}\)

Two identical capacitors \(C_{1}\) and \(C_{2}\) of equal capacitance are connected as shown in the circuit. Terminals \(a\) and \(b\) of the key \(k\) are connected to charge capacitor \(C_{1}\) using a battery of emf \(V\) volt. Now disconnecting \(a\) and \(b\) terminals, terminals \(b\) and \(c\) are connected. Due to this, what will be the percentage loss of energy?

Three identical capacitors are connected as follows:
 
Which of the following shows the order of increasing capacitance (smallest first)?

Two capacitors of capacitance \(6~\mu\text{F}\) and \(3~\mu\text{F}\) are connected in series with battery of \(30~\text{V}\). The charge on \(3~\mu\text{F}\) capacitor is:
          
1. \( 3 ~\mu\text{C}\)
2. \( 1.5 ~\mu\text{C}\)
3. \( 60~\mu\text{C}\)
4. \( 900~\mu\text{C}\)

The graph of electric potential \((V)\) versus distance \((r)\) is shown in the diagram.  The value of the electric field at a distance \(A\) will be:

The equivalent capacitance of the given circuit is:

        
1. \(\dfrac{C}{2}\)
2. \(C\)
3. \(2C\)
4. \(4C\)