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(A) When a dielectric material with dielectric constant $K > 1$ is inserted between the plates of a capacitor (assuming the capacitor is disconnected

Vedclass · Accepted Answer

Increase,decrease,decrease

Vedclass · Answer

(A) When a dielectric material with dielectric constant $K > 1$ is inserted between the plates of a capacitor (assuming the capacitor is disconnected from the battery,so charge $Q$ remains constant):$1$. Capacitance $(C)$: The new capacitance is $C' = KC$. Since $K > 1$,the capacitance increases.$2$. Potential Difference $(V)$: Since $V = Q/C$,and $C$ increases while $Q$ is constant,the potential difference $V$ decreases.$3$. Potential Energy $(U)$: The potential energy is given by $U = Q^2 / (2C)$. Since $C$ increases and $Q$ is constant,the potential energy $U$ decreases.Therefore,the correct sequence is: Increase,decrease,decrease.

When a dielectric substance is placed between the two plates of a capacitor,what happens to its capacitance,potential difference,and potential energy,respectively?

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