Question

QUISTION 1: Calculate the electrical field strength between a pair of plates spaced 10 mm apart when a p.d. of 0.5 kV exists between them. 10 The capacitance of a pair of plates is 40 pF when the dielectric between them is air. If a sheet of glass is placed between the plates (so that it completely fills the space between them), calculate the capacitance of the new arrangement if the relative permittivity of the glass is 6. A two plate capacitor has a value of 47 nF. If the plate area was doubled and the thickness of the dielectric was halved, what then would be the capacitance?

Answer 1

Ans dh 1. 3. Given Data de 10mm Pd = 0.5kV Electric field strength 0-5KY = AV = 0.5X103 Toxio3 = 5x104 V/m plate Capacitance formula 10. Parallel c = tot d E = permittivity between plate (Air) of any other medium - Ex = kEo where ere K= dielectric constant oF medium Now to A = 40PF when air d replaced by glass K=6 It is li', So 6EoA = 6X40 PF = 240PF New = kd

of previous problem This is the extension to A - 47nf đ Now According to question of plate Area doubled & thickness of dielectric is halved Anew=2A depend doesn't on thickness the of capacitance dielectric 2 2 EOA €o (2A) New = - 2x47nF CNew = 60 (20) = 2 ( 6 ) = L. = 94nF * Ang