An air-filled capacitor consists of two parallel plates, each with an area of 10

Question

An air-filled capacitor consists of two parallel plates, each with an area of 10.02 cm^2, separated by a distance of 1.04 mm. A 14 V potential difference is applied to these plates. Find the magnitude of the electric field between the plates. The permittivity of a vacuum is 8.8542 Times 10^-12 C^2/N middot m^2. Answer in units of V/m. 022 (part 2 of 4) 10.0 points Find the magnitude of the surface charge density. Answer in units of C/m^2. Find the capacitance. Answer in units of F. Find the magnitude of the charge on each plate. Answer in units of pC.

Explanation / Answer

electric field=V/d=14/(1.04*10^(-3))=13461.54 N/C

area=10.02*10^(-4) m^2

distance=1.04*10^(-3) m

capacitance=epsilon0*area/distance=8.53*10^(-12) F=8.53pF

charge=CV=119.42*10^(-12)=119.42 pC

surface charge density=charge/area=1.19*10^(-7) C/m^2

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