Question

Explain the principle of capacitor. Deduce an expression for the capacitance of the parallel plate capacitor.

Answer 1

Principle of Capacitor :-
It works on the principle , when the earthen conductor brought across to it, the capacitance of the conductor increase immediately. It separates with the two plates having equal and opposite charge on it.
 
Expression for capacitance of the parallel plate capacitor. 

Refers to the diagrammatic attachments :- The charges shown in the figure +Q & -Q always uniform , If the plates separation is almost small then electric field strength b/w plates is to be uniform. 

Assume A be the area, d is the distance & K is the dielectric constant here.
S. Charge density 
σ = $\frac{Q}{A}$ 
Now moving to electric field strength b/w the plates , then  
E = σ / Kε₀ 
Here ε₀ → permittivity in vacuum .

Now .Potential difference b/w plates.
$V_{AB} = E_{d}$  = σd / Kε₀ ----(i)
Now put the value of σ 
$V_{AB}$ = [(Q/A) × d ] /Kε₀ 

$V_{AB}$ = Qd / Kε₀.A
Now for the capacaitance C.
C =$\frac{Q}{V_{AB}}$ 
C = Q / [Qd / Kε₀.A]
Or ,
___________
C = Kε₀ .A / d
___________



Hope it Helps

Answer 2

Answer:

Refer to the attached images...