A capacitor is charged by using a battery,whi | vedclass

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Question

(C) When a capacitor is charged and the battery is disconnected,the charge $Q$ on the plates remains constant because there is no path for the charge

Vedclass · Accepted Answer

Decrease in the potential difference across the plates,reduction in stored energy,but no change in the charge on the plates.

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(C) When a capacitor is charged and the battery is disconnected,the charge $Q$ on the plates remains constant because there is no path for the charge to flow.When a dielectric slab of dielectric constant $K$ is inserted between the plates,the capacitance increases from $C$ to $C' = KC$.Since $Q = CV$,the new potential difference $V'$ is given by $V' = Q / C' = Q / (KC) = V / K$. Since $K > 1$,the potential difference decreases.The stored energy $U$ is given by $U = Q^2 / (2C)$. The new energy $U'$ is $U' = Q^2 / (2C') = Q^2 / (2KC) = U / K$. Since $K > 1$,the stored energy decreases.Therefore,the potential difference decreases,the stored energy decreases,and the charge remains constant.

List-$I$	List-$II$
$(A)$ $[\mathbf{a} \mathbf{b} \mathbf{c}]$	$1. \|\mathbf{a}\|\|\mathbf{b}\|\cos(\mathbf{a}, \mathbf{b})$
$(B)$ $(\mathbf{c} \times \mathbf{a}) \times \mathbf{b}$	$2. (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$
$(C)$ $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$	$3. \mathbf{a} \cdot \mathbf{b} \times \mathbf{c}$
$(D)$ $\mathbf{a} \cdot \mathbf{b}$	$4. \|\mathbf{a}\|\|\mathbf{b}\|$
	$5. (\mathbf{b} \cdot \mathbf{c})\mathbf{a} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$

$A$ capacitor is charged by using a battery,which is then disconnected. $A$ dielectric slab is then inserted between the plates. What is the result?

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