01---- A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field, and energy associated with this capacitor are given by Q0, V0, E0 and U0 respectively. A dielectric slab is now introduced to fill the space between the plates with battery still in connection. The corresponding quantities are now given by Q, V, E and U are related to previous quantities as (a) Q> Q0 (b) V> V0 (c) E> E0 (d) U> U0 
 
02---A parallel plate air capacitor with no dielectric between the plates is connected to the constant voltage source. How would capacitance and charge change if dielectric of dielectric constant K=2 is inserted between the plates. C0 and Q0 are the capacitance and charge of the capacitor before the introduction of the dielectric. (a) C=C0/2 ; Q=2Q0 (b) C=2C0 ; Q=Q0/2 (c) C=C0/2 ; Q=Q0/2 (d) C=2C0 ; Q=2Q0

Asked by shreeshampandey | 19th Nov, 2015, 11:50: AM

Expert Answer:

The potential difference 'V' between the plates remains unchanged (i.e., equal to the voltage of the battery) because the capacitor 'C' remains connected to the battery.

i.e., C = KC0

Thus, V = V0.

The introduction of the dielectric slab increases capacitance C. Hence, Q (= CV) increases.

Q = CV = (KC0)V0 = KQ0

As K>1, Q>Q0.

Since the plate separation ‘d’ remains unchanged, V remains unchanged and electric field, begin mathsize 14px style straight E equals straight V over straight d end styleremains unchanged.
begin mathsize 14px style therefore straight E equals straight V over straight d equals straight V subscript 0 over straight d equals straight E subscript 0 end style

i.e., E = E0

begin mathsize 14px style therefore end styleThe energy stored, begin mathsize 14px style straight U equals 1 half CV squared end style increases because 'C' increases.
begin mathsize 14px style straight U equals 1 half CV squared space space space equals 1 half left parenthesis KC subscript 0 right parenthesis left parenthesis straight V subscript 0 right parenthesis squared space space space equals KU subscript 0 end style
 
As, K>1, U>U0.
Hence, option (a) and (d) are correct.

Note: Kindly ask the other question as a separate query.

Answered by Faiza Lambe | 19th Nov, 2015, 02:06: PM

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