A 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
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(d) When there is no battery, charge remains same while potential difference and electric field decreases
i.e. $Q' = {Q_0},V' = \frac{{{V_0} \times 3}}{9} = \frac{{{V_0}}}{3}$and $E' = \frac{{{E_0} \times 3}}{9} = \frac{{{E_0}}}{3}$
i.e. $Q' = {Q_0},V' = \frac{{{V_0} \times 3}}{9} = \frac{{{V_0}}}{3}$and $E' = \frac{{{E_0} \times 3}}{9} = \frac{{{E_0}}}{3}$
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