Question

Two parallel plates of area 160 cm2 are given charges of equal magnitudes 9.5 × 10-7 C but opposite signs. The electric field within the dielectric material filling the space between the plates is 1.0 × 106 V/m. (a) Calculate the dielectric constant of the material. (b) Determine the magnitude of the charge induced on each dielectric surface. (a) Number Enter your answer for part (a) in accordance to the question statement

Answer 1

Answer: a) 6.7 ; b) -9.5 * 10^-7 C induced in the dielectric on the positive plate and 9.5 * 10^-7 C induced in the other side of the dielectric close to the negative plate.

Explanation: In order to solve this problem we have to use the formule for two parallel plates capacitor, which is given by:

C= εo*A/d where A and d are the cross section and separation of the plates, respectively.

Aditionally we know that C= Q/ΔV and ΔV = E*d

By uning both expression we have:

C= εo*A/d=Q/ΔV=εo*A/d=Q/E*d then

with a dielectric between the plates, the new C is obtained by multiply by k the initial capacitance.

finally we have

εo*k*A=Q/E

then k= Q/(E*εo*A)= 6.7

In the dielectric is induced oposite charge to each plate of the capacitor, The value of induced charge in the dielectric is the same for the capacitor but with oposite sign.