The parallel plates of a capacitor have an area 0.2 m 2 and are 10- 2 m apart. The original potential difference between them is 3000 V, and it decreases to 1000 V when a sheet of dielectric is inserted between the plates filling the full space. Compute: ( ∈ 0 = 9 x 10- 12 S. I. units)
(i) Original capacitance C 0 .
(ii) The charge Q on each plate.
(iii) Capacitance C after insertion of the dielectric.
(iv) Dielectric constant K.
(v) Permittivity ∈ of the dielectric.
(vi) The original field E 0 between the plates.
(vii) The electric field E after insertion of the dielectric.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
[(i) 20 ∈ 0 = 180 pF, (ii) 5.4 × 10 –7 C, (iii) 540 pF, (iv) 3, (v) 27 x 10 -12 C 2 N -1 m -2 , (vi) 3 × 10 5 V/m, (vii) 1 x 10 5 V/m]
Sol. (i) C 0 = 
= 
= 20 ∈ 0
= 20 × 9 × 10 –12 = 180 pF
(ii) Q = C 0 V = 180 × 10 –12 × 3000 = 5.4 × 10 –7 C
(iii) C 1 = 
= 
= 540 pF
(iv) K = 
= 
= 3
(v) ∈ = ∈ r ∈ 0 = K ∈ 0 = 3 × 9 × 10 –12 = 27 × 10 –12 C 2 N –1 m –2
(vi) E 0 = 
= 
= 3 × 10 5 V/m
(vii) E = 
= 
= 1 × 10 5 V/m.
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