The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm The capacitor is charged by connecting it to a 400 V supply

  1. How much electrostatic energy is stored by the capacitor?
  2. View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume u. Hence arrive at a relation between u and the magnitude of electric field E between the plates.

Asked by Topperlearning User | 22nd Apr, 2015, 09:59: AM

Expert Answer:

Area of the plates of a parallel plate capacitor, A = 90 cm2 = 90 x 10-4m2 Distance between the plates, d = 2.5 mm = 2.5 x 10-3 m potential difference across the plates, V = 400 V

  1. Capacitance of the capacitor is given by the relation,

           begin mathsize 11px style straight C equals fraction numerator element of subscript 0 straight A over denominator straight d end fraction end style

Electrostatic energy stored in the capacitor is given by the relation, begin mathsize 11px style straight E subscript 1 equals 1 half CV squared end style

             begin mathsize 11px style equals 1 half fraction numerator element of subscript 0 straight A over denominator straight d end fraction straight v squared end style

Where,

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '. = Permittivity of free space = 8.85 x 10-12C2N-1m-2

 begin mathsize 11px style straight E subscript 1 equals fraction numerator 1 cross times space 8.85 space cross times space 10 to the power of negative 12 end exponent space cross times space 90 space straight x 10 to the power of negative 4 end exponent space cross times space left parenthesis 400 right parenthesis squared over denominator 2 space cross times space 2.5 space cross times space 10 to the power of negative 3 end exponent end fraction equals space 2.55 space cross times space 10 to the power of negative 6 end exponent straight J end style

Hence, the electrostatic energy stored by the capacitor is 2.55 x 10-6J

(b) Volume of the given capacitor,

V' = A x d

    = 90 x 10-4 x 2.5 x 10-3

    = 2.25 x 10-5 m3

Energy stored in the capacitor per unit volume is given by,

begin mathsize 12px style u equals fraction numerator E subscript 1 over denominator V apostrophe end fraction
space space space equals fraction numerator 2.55 space x space 10 to the power of negative 6 end exponent over denominator 2.25 space x space 10 to the power of negative 4 end exponent end fraction equals 0.0113 space Jm to the power of negative 3 end exponent end style

Again, begin mathsize 11px style straight u fraction numerator straight E subscript 1 over denominator straight V apostrophe end fraction end style

begin mathsize 11px style equals fraction numerator begin display style 1 half CV squared end style over denominator Ad end fraction equals fraction numerator begin display style fraction numerator element of subscript 0 straight A over denominator 2 straight d end fraction end style over denominator Ad end fraction equals 1 half element of subscript 0 open parentheses straight V over straight d close parentheses squared end style

Where,

begin mathsize 11px style straight V over straight d end style = Electric intensity = E

begin mathsize 11px style therefore straight u equals space 1 half element of subscript 0 space straight E squared end style

(c) When zero of potential energy is taken, d1=1.06 Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Therefore, Potential energy of the system = Potential energy at d1 - Potential energy at d

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

= 21.73 x 10-19J - 27.2eV

= 13.58eV - 27.2eV

= -13.6eV

Answered by  | 22nd Apr, 2015, 11:59: AM