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The parallel plate of a capacitor have...

The parallel plate of a capacitor have an area `0.2 m^(2)` and are `10^(-2)` m apart. The original potential difference between them is 3000V, and it decreases to 1000 when a sheet of dielectric is inserted between the plates. Compute
(a) Original capacitance `C_(0)`
(b) The original charge Q on each plate .
(C) Capacitance C after insertion of the dielectric .
(d) Dielectric constant K .
(e) The original field `E_(0)` between the plates and .
(f) The electric field `E_(0)` between the plates and .
(f) The electric field E after insertion of the dielectric . `(epsi_(0) = 8.85 xx 10^(-12) S.I` unit).

Text Solution

Verified by Experts

(a) `C_(0) = (in_(0)A)/(d) = (8.85 xx 10^(-2) xx 0.2)/(0.01)F`
(b) `Q = C_(0) V = 1.77 xx 3000 C " " = 1.77xx 10^(-10)F` (c) As Q remains constant , So `C_(0) V_(0) = CV rArr C = (C_(0)V_(0))/(V) = (1.77xx 10^(-10) xx 1000)/(3000) = 5.90 xx 10^(-11)F`
(d) `K = (C)/(C_(0)) = (V_(0))/(V) = (3000)/(1000)= 3`
(e) `E_(0) = (V)/(d) = (3000)/(0.01) = 3xx 10^(5) V//m`
(f) ` E = (V)/(d) = (1000)/(0.01) = 10^(5)V//m`
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