Capacitance - Long Question Answer 6 | Physics Grade XI Solution

Capacitance Long Question Solution

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Capacitance - Long Question Answer 6 | Physics Grade XI Solution

Capacitance - Long Question Answer 6

Question: Obtain an expression of energy stored in a charged capacitor and use it to derive the expression of energy density of a dielectric medium in an electric field.

Answer: Energy stored in a charged capacitor: Suppose an uncharged capacitor of capacitance C is to be charged with charge Q till the potential difference becomes V, then
q = C V
 
When the capacitor is being charged let dq be charge at any instant and P.D. across the plate is V.
Then the work done by the charge on charging it is given by;
dW = V dq,
Since q = C V,
V = q/C,
SO, dW = q/C dq
 
Integrating both sides from 0 to q as limits,
  qʃ0dW = C qʃq dq
  or, W = C [q2/2]0q
  or, W = q2/2C
 
This work done is stored as (electrical potential) energy on the capacitor. Thus, energy stored in the capacitor is given by;
  E = q2/ 2C
Moreover, using q = C V, we get
  E = (CV)2 / 2C = ½ CV2
and
  E = ( ½ q)/ (q/V) 
  E = ½ q V
Hence, E = ½ qV = ½ (CV)2 =q2/ 2C

This is the required expression of energy stored by capacitor.

The energy stored per unit volume of a capacitor is called energy density.

i.e. Energy density, u = Energy stored/ Volume = CV2 / 2Ad
 = Є0 A/d * V2/2Ad
= Є0 A/2 * V2/ d2
= ½ Є0E2 [.: V/d]
This is energy density

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