**(a).The plates have been changed with Q and -Q ,**

By connecting them to the two terminals of a battery.

**Q is called the charge of the capacitor, **

Through the total charge of the capacitor is zero.

**Intervening medium :** air/vaccum

**Let us consider insulated metal plate A, **

Charged to Q, till its potential becomes maximum. Bring another insulated metal plate B near to plate A. By induction, a negative charge is produced on inner surface of B and equal positive charge on farther face as shown in figure.

The induced negative charge tends to decrease the potential of A and induced positive charge tends to increase the potential of A.

The overall potential of A reduces and hence some more charge can be given to A to raise its potential to the same maximum value.

**Thus,**

Capacity of conductor A has increased, by bringing another uncharged conductor B in its vicinity. Earthing B, makes positive charge flew to earth.

**Thus, **

Due to the induced negative charge on B, potential of A is greatly reduced.

**Thus, **

A large amount of charge can be given to A to raise it to the maximum potential.

**(b) Charge on the capacitor(single), **

Q = CV,

Where C is the capacitance of both the capacitors(common).

**∴ Energy stored in the single capacitor,**

E_{1} =** ****\(\frac{1}{2}cv^2\)**

= **\(\frac{1}{2}Qv\)**

**Now,**

**In 2nd case,**

When another uncharged capacitor of capacitance C, is connected to the single capacitor, then they reach common potential, V'.

**∴ Now, **

**The charge on each capacitor, **

Q' = CV'

**By charge conservation,**

Q' = \(\frac{Q}{2}\)** **

⇒ V' = \(\frac{V}{2}\)** **

**∴ Energy sotred in the combination, **

E_{1} =** ****\(\frac{1}{2}\times\)** 2** **\(\times\) Q' \(\times\) V'

= \(\frac{Q}{2}\)\(\times\)\(\frac{V}{2}\)

= **\(\frac{1}{4}\)**QV

**(c) **

**For the charge +q at A to be in equilibrium,**

Force on A due to Q at B+ Force on A due to Q at O = 0

\(\frac{1}{4\pi ∈_0}.\frac{q^2}{Z^2}\)

= \(\frac{1}{4\pi ∈_0}.\frac{q\times Q}{1^2}\)

Q = \(\frac{q^2}{4q}\)

= \(\frac{q}{4}\)