IP A parallel-plate capacitor filled with air has plates of area 6.1 times 10^-3

Question

IP A parallel-plate capacitor filled with air has plates of area 6.1 times 10^-3 m^2 and a separation of 0.31 mm. Find the magnitude of the charge on each plate when the capacitor is connected to a 11-V battery. Express your answer using two significant figures. Will your answer to part A increase, decrease, or stay the same if the separation between the plates is increased? Increase Decrease Stay the same Calculate the magnitude of the charge on the plates if the separation is 0.90 mm. Express your answer using two significant figures.

Explanation / Answer

Capacitance =Epsilon*A/d

= 8.84 x 10^-12 * 6.1*10^-3/(0.31*10^-3)

=1.7395*10^-10 F

Q=C*V

=1.7395*10^-10 *11

=1.9134 nC

Charge on plates =+1.9134 nC and -1.9134 nC

b) if separation increases then Capacitance decreases and hence charge decreases.

c)Cnew =0.599*10^-10 F

Charge =Cnew*V

=6.591*10^-10 C

=0.659 nC

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