The figure shows a parallel-plate capacitor of plate area A and plate separation

Question

The figure shows a parallel-plate capacitor of plate area A and plate separation d. A potential difference V_0 is applied between the plates. While the battery remains connected, a dielectric slab of thickness b and dielectric constant k is placed between the plates as shown. Assume A = 155 cm^2, d = 1.27 cm, V_0 = 59.2 V, b = 0.764 cm, and kappa = 4.11. Calculate the capacitance, the charge on the capacitor plates, the electric field in the gap, and the electric field in the slab, after the slab is in place.

Explanation / Answer

a)

The capacitance of the capacitor is given by

C =Aeo/d-b+(b/k) =(155*10-2m2)(8.85*10-12C2/Nm2)/[1.27*10-2-0.764*10-2+(0.764*10-2/4.11)]

   =1371.75*10-14/(5.06*10-3 +1.858*10-3) =1371.75*10-14/6.918*10-3 =198.28*10-11F =1982.8pF

b)

Now the charge on the capacitor plates is

Q =CV =(198.28*10-11F)(59.2V) =1.1738*10-7C

c)

The electric field in the slab is given by

E =Q/Aeo =1.1738*10-7C/(155*10-2m2)(8.85*10-12C2/Nm2)=8.55*103V/m or N/C

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