For a point \( Q \) on the perpendicular bisector of an electric dipole (distance \( r \) from the center, where \( r \gg \text{dipole length} \)):

1. Electric Field on Perpendicular Bisector: The electric field \( E \) at a point on the perpendicular bisector of a dipole is given by:
\[
E \propto \frac{p}{r^3}
\]

2. Dependence:
- Directly proportional to the dipole moment \( p \).
- Inversely proportional to \( r^3 \).

Answer:
The electric field at \( Q \) is proportional to:
\[
p \quad \text{and} \quad r^{-3}
\]

Number of Electric field line in space represents electric field intensity. As number of field line is maximum at A followed by B then at C. so, \(E_{A}>E_{B}>E_{C}\)

By Gauss's law, the total flux through the cube is \(\Phi = \frac{q_{\text{enclosed}}}{ε_0}\). Since the charge is at the center, the flux through one of the six faces is \(\Phi_1 = \frac{\Phi}{6} = \frac{Q \times 10^{-6}}{6ε_0}\).

According to Gauss's Law, total flux through the cube is \(\phi = \frac{q}{\epsilon_0}\). Since a cube has 6 identical faces, the flux through one face is \(\phi' = \frac{\phi}{6} = \frac{Q \times 10^{-6}}{6\epsilon_0}\).

The potential of a coalesced drop is related to the individual potential by \(V' = n^{2/3} V\). For \(n = 1000\), \(V' = (1000)^{2/3} \times 1 = 100 \text{V}\).

Torque is \(\tau = pE \sin 30^\circ = pE/2 \implies pE = 2\tau\). Work done is \(W = -pE(\cos 60^\circ - \cos 30^\circ) = pE(\cos 30^\circ - \cos 60^\circ) = 2\tau(\frac{\sqrt{3}}{2} - \frac{1}{2}) = (\sqrt{3}-1)\tau\).

Work done by the electric field \(\vec{E} = E\hat{i}\) is given by \(W = q\vec{E} \cdot \vec{d} = qE\hat{i} \cdot (a\hat{i} + b\hat{j}) = qEa\). By work-energy theorem, \(K_f - K_i = W \implies K_f = qEa\).

The potential of a sphere with surface charge density \( \sigma \) is \( V = \frac{\sigma R}{\varepsilon_0} \). Since \( R_1 > R_2 \), \( V_1 > V_2 \). Charge flows from higher potential (larger sphere) to lower potential (smaller sphere).