(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,
E1 = \(\frac{1}{2}cv^2\)
= \(\frac{1}{2}Qv\)
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,
E1 = \(\frac{1}{2}\times\) 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 = \(\frac{q^2}{4q}\)
= \(\frac{q}{4}\)