A capacitor has a capacitance of 15 nF. Then a dielectric material is added in b

Question

A capacitor has a capacitance of 15 nF. Then a dielectric material is added in between the plates with k=3.0. What is the new capacitance of the capacitor (with dielectric)? nF How much charge will the new capacitor store on each plate when 200 Volts are put across it? mu C At a certain point in time, the capacitor is discharged; the two sides are connected to each other by something such as a wire, allowing the charges on one side to flow to the other. If it takes 2.0 seconds for all the positive and negative charges to equalize, what was the average electric current during this time? mu A

Explanation / Answer

New capacitance = k * C

                            = 3 * 15

                             = 45 nF

charge new capacitor store = CV

                                           = 45 * 10-9 * 200

                                           = 9 microC

average electric current = Q/t

                                     =   9/2

                                     = 4.5 microA

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