An air-filled capacitor consists of two parallel plates, each with an area of 7.

Question

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm^2 separated by a distance of 2.00 mm. A 19.0 V potential difference is applied to these plates. Calculate the electric field between the plates. 95 Remember that the electric field is uniform between the plates. What is the relationship between the electric field and electric potential in this case? kV/m Calculate the surface charge density. _____ nC/m^2 Calculate the capacitance. _________ pF Calculate the charge on each plate. _________ pC

Explanation / Answer

a) E=V/d

V=19 volts d=2mm =2*10-3m

E=19/(2*10-3m) =9.5kV/m

b) surface charge density= E*e0

E = electric field

e0= 8.852*10-12 F/m

surface charge density= =9.5kV/m *   8.852*10-12 F/m

=84.094 nC/m

C) q= surface charge density*area

=84.094 nC/m *2*10-3m

=168.188 pC

Q= CV

C= Q/V

C=168.188 pC/19V

C=8.851pF

d) Q=168.188 pC

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