Capacitance PBQ - Circuits - Kaplan

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trs494

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I came across this question in the PBQ's for chapter 6 and thought the explanation was incorrect for the answer given...

Shouldn't it be that doubling the area and inserting a dielectric material between the capacitor both double capacitance? Therefore, for the amount of time to charge the capacitor would double for roman numerals II and III?

My logic was just using C = C(initial) * k along with C = Q/V where voltage is constant to find when charge would double.

Images are attached below! Thanks!

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