A parallel-plate capacitor is constructed from two circular plates with diameter

Question

A parallel-plate capacitor is constructed from two circular plates with diameter 15.0 cm separated by 0.70 mm. What charge will collect on the plates if the capacitor is connected to a potential difference of 66.0 V?

The capacitor from the previous problem is now disconnected from the voltage source. The separation between the capacitor plates is quadrupled and an insulator is inserted between the plates. The voltage between the plates is measured to be 14.6 V. Find the dielectric constant of the insulator.

Explanation / Answer

a)Given that diameter is D=15cm so the radius is r=D/2=7.5cm=7.5 x 10-2m seperation is d=0.70 x 10-3 m the capacitance is C=or2/d                                =0.223 x 10-9 C potential is V=66V so the charge is Q=CV=14.74 nC b)since the battery is disconnected sme charge will remain and the potential is given by V1=14.6V so the capacitance is C1=Q/V1                                    =1.0099 x 10-9 F since the distance is quadrapuled new capacitance C1=kor2/4d                                 =k/4(C)                            k=4C1/C                             =18.11 b)since the battery is disconnected sme charge will remain and the potential is given by V1=14.6V so the capacitance is C1=Q/V1                                    =1.0099 x 10-9 F since the distance is quadrapuled new capacitance C1=kor2/4d                                 =k/4(C)                            k=4C1/C                             =18.11

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