1. Tardigrade
  2. Question
  3. Physics
  4. 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 the battery still in connection. The corresponding quantities now given by Q, V, E and U are related to the previous one as

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Q. 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_0,V_0,E_0 \, and \, U_0 $ 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

IIT JEEIIT JEE 1985Electrostatic Potential and Capacitance

Solution:

When dielectric slab is introduced capacity gets increased
while potential difference remains unchanged.
$\therefore V=V_0, C>C_0$
$Q=CV \, \, \, \, \, \, \, \, \therefore \, \, \, \, Q>Q_0 $
$U= \frac {1}{2}CV^2 \, \, \, \, \, \therefore \, \, \, \, \, U>U_0 $
$E= \frac {V}{d} $ but V and d both are unchanged
Therefore, $\, \, \, \, \, \, \, \, E=E_0$
Therefore, correct options are (a) and (d).