Physics, asked by xcreator52, 10 months ago

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 Q., V., E, and U.
respectively. A dielectric slab is now introduced to fill the space between the plates with the
battery still in connection. The corresponding quantities now given by Q, V, E and U are
related to the previous one as:

Answers

Answered by nehasarajohn

Hope it helps and please mark me as BRAINLIEST.........

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Answered by ratanvoleti

Answer:

Explanation:

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 previousquantities 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

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