An air-filled parallel plate capacitor has circular plates with radius r=20.0 cm, separated distance 4.00 mm. The capacitor is connected to 70.0 V battery. a) Find the capacitance, Co, and the charge on the plates, Qo. b) Find the electric field strength between the plates. c) While connected to the battery, a dielectric sheet with dielectric constant k is inserted between the plates (it fills the entire space). What are the capacitance, the charge, and the voltage now?

Answer: a) 278 * 10^-12 F and 19.4 * 10^-9 C

b) 17.44 * 10^3 N/C and c) C=k*C0 and V=70/k

Explanation: In order to solve this problem we have to use the expression of the capacitor of parallel plates as:

C=A*ε0/d where A is the area of the plates and d the distance between them

C=Π r^2*ε0/d

C=π*0.2^2*8.85*10^-12/0.004=278 * 10^-12F = 278 pF

then

ΔV = Q/C

so Q = ΔV*C=70V*278 pF=19.4 nC

The electric field between the plates is given by:

E = Q / (A*ε0) = 19.4 nC / (π*0.2^2*8.85*10^-12) = 17.44 * 10^3 N/C

If it is introduced a dielectric between the plates, then the new C is increased a factor k while the potential between the plates decreases a factor 1/k.