A) what is the potential difference between the plates? v= b)what is the area of

Question

A) what is the potential difference between the plates? v= b)what is the area of each plate? m^2 c.) what is the electric field magnitude between plates? =V/m d.)what is the surface charge density on each plate? C/m^2 A parallel-plate air capacitor has a capacitance of 600 pF and a charge of magnitude 0.200 uC on each plate. The plates are 0.600 mm apart. what is the potential difference between the plates? v= what is the area of each plate? m2 what is the electric field magnitude between plates? =V/m what is the surface charge density on each plate? C/m2

Explanation / Answer

capaciatcne C = eoA/d

where A = area , d= distance between the palte and e0 = constant = 8.85*10^-12

a,so charge Q is related to capacitance C as Q = CV

where is voltage

so now

a. V = Q/C =0.2*10^-6/(600*10^-12) = 333.33 volts

c, Eletric field E = V/d = 333.33/0.6 mm = 5.55*10^5 V/m

d. surface charge density = charge/area = 0.2*10^-6/0.0406 = 4.92 *10^-6 C/m^2

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