A lecturer wishing to demonstrate the effects of dielectrics does the following.

Question

A lecturer wishing to demonstrate the effects of dielectrics does the following. He connects a parallel plate capacitor to a battery and charges it to a charge Q. He then begins to insert a dielectric into the capacitor, and when it is half way in he remembers that he wanted to disconnect the battery. He does so (the dielectric is half way in when the battery is disconnected) and then inserts the dielectric the rest of the way. Which one of the following statements is true about what happens to the charge Q on the capacitor plates and the voltage difference V between the plates from just before when the dielectric was inserted to after it was fully inserted?

A. Q increases; V increases B. Q increases; V stays the same C. Q increases; V decreases D. Q stays the same; V increases E. Q stays the same; V stays the same F.  Q stays the same; V decreases G. Q decreases; V increases H. Q decreases; V stays the same I. Q decreases; V decreases

Explanation / Answer

answer is C.

as when dielectric constant increases so the Capacitance of the parallel plate capacitor increases.As when the Circuit is disconnected charge on the capacitor increases so as we all know Q=CV

so that Voltage also have to decrease to counterbalance the increase in capacitance.If the capacitor is no longer connected to the supply when you insert the dielectric then as the charge cannot change ( charge is conserved) the voltage will drop.And when it is connected Charge must increase to keep the potential constant.

Answer C=Q increases; V decreases

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