As per the given question,
Plate area of the capacitor (A)=3.0cm2=3.0×10−4m2
Distance between the plates(d)=1mm=1×10−3m
Permittivity of free space (ϵo)=8.85×10−12F/m
i)
We know that the capacitance of the capacitor =dϵoA
Now, substituting the values,
C=1×10−38.85×10−12×3×10−4F
⇒C=26.46×10−13F
⇒C=2.646×10−12F
ii) V=6.0 Volt
Hence charge stored in the capacitor (q)=cV=2.646×10−12×6=15.876×10−12C
iii) Charge density on the positive plate of the capacitor =q/A
Substituting the values,
=3.0×10−415.876×10−12C/m2
=5.292×10−8C/m2
iv) Magnitude of the electric field between the plates E=d−ΔV
=3×10−46N/C=2×104N/C
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