When a capacitor is filled with a dielectric constant $K = 3$,the charge is $Q_0$,voltage is $V_0$,and electric field is $E_0$. If the capacitor is now filled with a dielectric constant $K = 9$,what will be the new charge,voltage,and electric field respectively?
- A$3Q_0, 3V_0, 3E_0$
- B$Q_0, 3V_0, 3E_0$
- C$Q_0, V_0/3, 3E_0$
- D$Q_0, V_0/3, E_0/3$
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