Answer :

(a)

Suppose an uncharged capacitor of capacitance is connected in series with a switch and a battery of emf . The switch is initially open. Then, the switch is closed at . Initially, the capacitor acts as a short circuit. The current flows and deposits charge in the capacitor. The plate with the positive terminal of the battery loose electron and the other plate gains electron. Due to this build-up of charges, potential difference across the capacitor build up. When this potential difference is equal to the emf of the battery, the capacitor is said to be fully charged and the capacitor acts like an open switch.

At time t, the capacitor has charge . Let the potential difference across the plates be given by:

A small charge is being transferred from one plate to another. The work in this process is given by:

Integrating, we get

Using ,

This work done is stored as energy on the capacitor.

(b)

Let the capacitance of the capacitor be .

The energy stored by the capacitor when it is charged is given by:

The charge on the capacitor is

Now, when the capacitor is connected to the other capacitor, they both get the same charge in equilibrium by symmetry. As the total charge remains the same, the charge on each capacitor is

Now, the energy stored is

The ratio is:

OR

Let charges and be separated by a distance . Let point P be any point on the equatorial line of the dipole. Let it be a distance form the line joining the two charges.

Now,

The electric field at P due to the charges is:

The vertical components will cancel out. We need to add the horizontal components. Hence,

Let be the unit vector in the direction of the dipole moment (from positive charge to negative charge).

Now, by trigonometry,

If ,

(b)

All charges must be in equilibrium. As the coulomb force depends upon distance and by symmetry, the charge Q must be in the middle of the two charges. For the forces to balance on the q’s, Q must have opposite sign to that of q.

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