Answer to Question #125891 in Electric Circuits for Meet Darji

Question #125891
A parallel-plate capacitor has an area A = 3.00 cm2
and a plate separation d =1 mm. (a) Find
its capacitance. (b) How much charge is on the positive plate if the capacitor is connected to a
6.00-V battery? Calculate (c) the charge density on the positive plate, assuming the density is
uniform, and (d) the magnitude of the electric field between the plates
1
Expert's answer
2020-07-10T10:33:46-0400

As per the given question,

Plate area of the capacitor (A)=3.0cm2=3.0×104m2(A)=3.0 cm^2=3.0\times 10^{-4}m^2

Distance between the plates(d)=1mm=1×103m(d)= 1mm=1\times 10^{-3}m

Permittivity of free space (ϵo)=8.85×1012F/m(\epsilon_o) =8.85\times 10^{-12}F/m

i)

We know that the capacitance of the capacitor =ϵoAd=\frac{\epsilon_o A}{d}

Now, substituting the values,

C=8.85×1012×3×1041×103FC=\frac{8.85\times 10^{-12}\times 3\times 10^{-4}}{1\times 10^{-3}}F

C=26.46×1013F\Rightarrow C=26.46\times 10^{-13}F

C=2.646×1012F\Rightarrow C= 2.646\times 10^{-12}F


ii) V=6.0 Volt

Hence charge stored in the capacitor (q)=cV=2.646×1012×6=15.876×1012C(q)= cV =2.646\times 10^{-12}\times6 =15.876\times 10^{-12}C


iii) Charge density on the positive plate of the capacitor =q/A=q/A

Substituting the values,


=15.876×10123.0×104C/m2=\frac{15.876\times 10^{-12}}{3.0\times 10^{-4}}C/m^2


=5.292×108C/m2=5.292\times 10^{-8}C/m^2


iv) Magnitude of the electric field between the plates E=ΔVdE=\frac{-\Delta V}{d}

=63×104N/C=2×104N/C=\frac{6}{3\times 10^{-4}}N/C =2\times 10^{4}N/C


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