CBSE Class 12-science Answered
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 | 19 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, 
remains unchanged.
remains unchanged.
i.e., E = E0
The energy stored, 
increases because 'C' increases.
As, K>1, U>U0.
Hence, option (a) and (d) are correct.
Hence, option (a) and (d) are correct.
Note: Kindly ask the other question as a separate query.
Answered by Faiza Lambe | 19 Nov, 2015, 02:06: PM
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