Constants Part A A parallel plate capacitor has capacitance = 7.80 pF when there

Question

Constants Part A A parallel plate capacitor has capacitance = 7.80 pF when there is air between the plates. The separation between the plates is 1.70 mm What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00x104 V/m? Express your answer with the appropriate units Q- 398 pO Correct Part B A dielectric with K -2.60 is inserted between the plates of the capacitor, completely filling the volume between the plates. Now what is the maximum magnitude of charge on each plate if the electric field between the plates is not to exceed 3.00*104 V/m Express your answer with the appropriate units. e- Submit X Incorrect; Try Again; 2 attempts remaining

Explanation / Answer

According to the concept of the electric potential and capacitance

Charge Q=VCo

Given that

Capacitance Co=7.8pF

Electric field E=3*10^4 v/m

Distance d=1.7*10^-3 m

Now we find the maximum magnitude of the charge

Charge Q=(Ed) *Co

=(3*10^4*1.7*10^-3)*7.8*10^-12

=398pC

Now we find the charge after dielectric placed b/w plates

Capacitance C=kCo=2.6*7.8*10^-12=20.3pF

Charge Q=(Ed) C

=(3*10^4*1.7*10^-3)*20.3*10^-12

=1034.3 pF

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