If \theta is angle between \( \overrightarrow { P } \ and\ \overrightarrow { E } , then \theta =0^o\) for stable equilibrium, \( \theta=180^o\) for unstable equilibrium.

Torque tries to align the dipole along the field.

Zero

Zero. As \(W = q\Delta V= 500\mu C \times 0 = 0 \)

Potential at the middle point is zero. The values of potential at the two ends of the conductor must be +110V and -110V, So that potential difference across the two ends is 110-(-110) = 220V.

1F = 10¹²pF

No, the maximum charge that can be given to a capacitor is determined by the capacity of the condenser.

This is because density of air in ionosphere is so low that very few molecules only bombard the spaceship.

Let r be the radius of each small drop and R be the radius of bigger drop.
As volume of bigger drop = volume of n small drops
\({4\over 3}\pi R^3=n\times{4\over 3}r^3\)
\(R=n^{1/3}r\)
Also, potential of bigger drop = \(total \ charge\over capacity\)
\(V={nq\over 4\pi\epsilon_o R}={nq\over 4\pi\epsilon_on^{1/3}r}=n^{2/3}{q\over 4\pi\epsilon_o r}\)
\(=n^{2/3}\) times the potential of each small drop.

On introducing a thin metal oil, d is halved. Arrangement is equivalent to two condensor each of capacity 2 C in series.
\({1\over C_s}={1\over 2C}={1\over 2C}={1\over C}\)
\(C_s=C\)
Capacity is unchanged.

Dielectric constant varies inversely with temperature, i.e., dielectric constant decreases with rise in temperature.

We know that a positively charged particle is attracted towards the negatively charged plate and a negatively charged plate and a negatively charged particle is attracted towards the positively charged plate.
Here particle 1 and 2 are attraced positive that mean 1 and 2 are negatively charged. 3 is attracted towards negatively charged plate so it is positively charged.
As the deflection in the path of a charged particle is directly proportional to the charge/mass ratio
 \(y\propto \ \frac { q }{ m } \)
Here, the deflection in particle 3 is maximum. so,  the charge to mass ratio of particle 3 is maximum.