Q36 / 50 · Manipal (MET) 2023 · Physics

A charged capacitor when filled with a dielectric $K=3$ has charge $Q_0$, voltage $V_0$ and field $E_0$. If the dielectric is replaced with another one having $K=9$, the new values of charge voltage and field will be respectively

  • A
    $3 Q_0, 3 V_0, 3 E_0$
  • B
    $Q_0, 3 V_0, 3 E_0$
  • C
    $Q_0, \frac{V_0}{3}, 3 E_0$
  • Correct answer
    D
    $Q_0, \frac{V_0}{3}, \frac{E_0}{3}$

Solution

When charging battery is disconnected then charge remains constant, $V$ becomes $1 / K$ times and $E$ becomes $1 / K$ times. So, new value of charge is $Q_0$. New value of $V=\frac{V_0 \times 3}{9}=\frac{V_0}{3}$ New value of $E=\frac{E_0 \times 3}{9}=\frac{E_0}{3}$

Hence, the correct answer is option: 'D'.

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