Question

A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric filed and energy associated with this capacitor are gives 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 are now given by $Q$, $V$, $E$ and $U$ are related to the previous one as:

• A. $Q>Q_0$
• B. $V>V_0$
• C. $E>E_0$
• D. $U>U_0$

Answer 1

The correct option is A $Q>Q_0$

$Q_0=C_0{V}_0$
$U_0=\frac{1}{2}{C}_0{V}_{)}^2$
$E_0=\frac{V_0}{\left(d\right)}$ (d is distance between plates).
when dielectric is added.
$c=k{C}_0$
As battery is still connected $U_0$ remains same.
$V=U_0$
$U=\frac{1}{2}\left(k{C}_0\right)V_0^2$
$U-k\frac{1}{2}{C}_0{V}_0^2$
$U=k{V}_0$
$Q=k{C}_0\cdot V_0.$
$\therefore Q=k{Q}_0$
$E=\frac{U}{d}$
$E=\frac{U_0}{d}$
$E=E_0.$