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(D) When a dielectric of constant $K$ is inserted while the voltage source remains connected,the potential difference $V$ remains constant.$1$. Capaci

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All of the above.

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(D) When a dielectric of constant $K$ is inserted while the voltage source remains connected,the potential difference $V$ remains constant.$1$. Capacitance: The new capacitance becomes $C' = KC$.$2$. Charge: The new charge $Q' = C'V = (KC)V = KQ$. Thus,the charge increases to $K$-times.$3$. Energy: The initial energy $U_i = \frac{1}{2}CV^2$. The final energy $U_f = \frac{1}{2}C'V^2 = \frac{1}{2}(KC)V^2 = KU_i$. Thus,the energy increases to $K$-times.$4$. Force: The force of attraction between the plates is given by $F = \frac{Q^2}{2A\epsilon_0 K_{dielectric}}$. Since the capacitor is filled with a dielectric,the effective force is $F = \frac{Q^2}{2A\epsilon_0 K}$. However,considering the force between the plates due to the field $E = V/d$,the force $F = QE/2$. Since $Q$ becomes $KQ$ and $E$ remains $V/d$,the force becomes $F' = (KQ)(V/d)/2 = K(QV/2d) = KF$. Wait,if we consider the force between the plates as $F = \frac{Q^2}{2A\epsilon_0}$,and $Q$ becomes $KQ$,then $F' = \frac{(KQ)^2}{2A\epsilon_0} = K^2 F$. Therefore,all statements are correct.

The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now,a dielectric of dielectric constant $K$ is inserted to fill the whole space between the plates,with the voltage source remaining connected to the capacitor.

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