A parallel-plate capacitor C=245 µF has a charge Q= 980 µC. The plates are 0.328 mm apart.?

a- What is the potential difference between the plates?  

b- Find the area of each plate. 

c- What is the electric field between the plates?

d- Calculate the surface charge density (σ) on each plate.

billrussell422021-03-04T12:52:28Z

Q = CV
V = Q/C = 980 µC / 245 µF = 4 volts

Parallel plate cap
C = ε₀εᵣ(A/d) in Farads
   ε₀ is vacuum permittivity, 8.854e-12 F/m
   εᵣ is dielectric constant or relative permittivity
    of the material (vacuum = 1)
   A and d are area of plate in m² and separation in m

245 µF  = (8.854e-12)(A) / (0.000328)
A = 9076 m²

E = V/d = 4 / 0.000328 = 12200 V/m

σ = Q/A = 980 µC / 9076 = 0.108 µC/m² or 108 nC/m²