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 now given by Q,V,E and U are related to the previous quantities as :
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a
Q>Q0
b
V>V0
c
E>E0
d
U
answer is A.
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Detailed Solution
∵Q=CVWith the introduction of dielectric slab, C increases but V remains unchanged because battery is connected.∴Charge (Q) increases due to increase in C,∴Q>Q0Also, U=12CV2∵C>C0,V=V0Then, U>U0