The plates of a parallel-plate capacitor each have an area of 0.1 m^2 and are separated by a 0.9 mm thick layer of porcelain. The capacitor is connected to a 14 V battery. (The dielectric constant for porcelain is 7.) (a) Find the capacitance. (b) Find the charge stored. (c) Find the electric field between the plates.

Answer: a) 6.88 nF; b) 96.32 nC; c) 108.83 * 10^3 N/C

Explanation: In order to solve this problem we need to used the expresion for the capacitor of two parallel plates, which is given by:

C=εo*A/d; A is the area and d the separaction between the plates.

inside the capacitor there is a porcelain layer so we have to multiply by the dielectric constant (k=7).

Then C=8.85*10^-12*7*0.1 / 0.0009=6.88 nF

Also we know that ΔV=Q/C

Q = C*ΔV = 6.88 nF * 14 V = 96.32 nC

Finally, The electric field between the plates is given by:

E = σ/εo = Q / (A*εo) = 96.32 / (0.1*8,85 * 10^-12) = 108.83 * 10^3 N/C