Question

two parallel plates (each of area 400 cm^2)of a capacitor are given equal and opposite charges of magnitude 3.56 micro coloumb. The space between the plates is filled with a dielectric. If the electric field in the dielectric is 1.2 × 10 ^ 6 V/ M. what is the dielectric constant of the material.

Answer 1

E=Q/(€o A K)

1.2*10^6 = 3.56*10^-6 /(8.85*10^-12*0.04* K)

K=8.83

Answer 2

Hey dear,

◆ Answer-
k = 8.38

◆ Explaination-
# Given-
Q = 3.56×10^-6 C
E = 1.2×10^6 V/m
A = 400 cm^2 = 0.04 m^2
ε = 8.85×10^-12

# Solution-
Electric field of the dielectric is given by-
E = Q / (εAk)
k = Q / (εAE)
k = (3.56×10^-6) / (8.85×10^-12 × 0.04 × 1.2×10^6)
k = 8.38

Therefore, dielectric constant of the medium is 8.38

Hope it helps...