Before connecting the circuit shown in figure all capacitors were uncharged. The circuit now is in steady state. Now the switch $S$  is closed. During the time the steady state is reached again:

Column?I	Column?II
i. Charge on C1	p. increases
ii. Charge on C2	q. decreases
iii. Charge on C3	r. remains same
iv. Charge on C4	s. becomes zero

Now, match the given columns and select the correct option from the codes given below.

Codes:

 

Not really. The electric dipole moment vector directs or points from the negative charge to the positive charge. But the electric field lines that a dipole creates will point away from the positive and move to the negative charge.

Yes, the cube, which is a closed surface containing only one electric dipole will make electric flux zero. This follows Gauss's Law when the total charge inside it is zero. The field lines entering the surface will exit, and that would result in zero net flux. 

The magnitude of each charge and the distance that separates them. 

Gauss Law is only concerned with the total enclosed charge that finally tells us the total flux. The charges outside may change field patterns. They not affect the total flux. It's actually incorrect to assume the field due to the external charges should also affect the flux through the Gaussian sur

Gauss Law does not directly give the electric field in all cases. It can only be used in calculations for symmetrical surfaces: spherical, cylindrical, or planar.