A capacitor is charged. Then the connections with the battery are removed. Next,

Question

A capacitor is charged. Then the connections with the battery are removed. Next, a piece of glass is shoved into the gap. How will the charge change? And the voltage? And the capacitance?
Same as previous questions, but now the connections with the battery are not removed? A capacitor is charged. Then the connections with the battery are removed. Next, a piece of glass is shoved into the gap. How will the charge change? And the voltage? And the capacitance?
Same as previous questions, but now the connections with the battery are not removed?
Same as previous questions, but now the connections with the battery are not removed?

Explanation / Answer

battery is removed.

So charge on capacitor will remain same.

capacitor will increase. [ C = e0 er A / d ]

( dielectric constant times)

V = Q / C => voltage will decrease.

When battery is connected.

then voltag drop will remain same.

Capacitance increase.

so charge increases. ( Q = C V )

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