Two parallel plates have equal and opposite charges. When the space between the plates is evacuted, the electric field is E_0=3.20xx10^5V//m. When the space is filled with electric the electric field is E-2.50xx10^5V//m a. What is the dielectric constant? b. What is the charge density on each surface of the dielectric?
Answers
When the space between the plates is evacuated, the electric field is E=Aϵ0Q where A be the area of plates.
When the space is filled with dielectric, the electric field is E=AKϵ0Q where K is the dielectric constant.
EE0=AKϵ0QAϵ0Q=K
∴K=EE0=2.50×1053.20×105=1.28≈1.3=13/10
GIVEN:
Two parallel plates, when the space between the plates is evacuated, E=3.20x10^5V/m.
When the space is filled with dielectric E-2.50xx10^5V/m
TO FIND:
a .The dielectric constant
b. What is the charge density on each surface of the dielectric?
SOLUTION:
◆We know,
The electric field between two plates when the space between the plates is evacuated,
◆E=AϵQ
= 3.20xx10^5V/m
◆where A is the area of plates.
ϵ - permittivity of the medium,
Q - the charge.
◆When the space is filled with dielectric,
Electric field ,E'=AϵQ /K
=2.50x10^5V/m
where K is the dielectric constant.
◆E/E' = (AKϵQ)/(AϵQ) = K
◆1/K=E'/E
=3.20×10^5/2.50×10^5
◆K=1.28.
◆Now,
Charge density,
D = ϵE
= 8.854× 10^-12 × 3.2×10^5
=2.833×10^-6
◆Charge density on dielectric,
D' = D (1-1/K)
= 2.83 ×10^-6 (1 - 1/1.28)
=6.2 × 10^-5 C/m^2.
ANSWER :
DIELECTRIC CONSTANT ,K = 0.7.
CHARGE DENSITY = 0.62 × 10^-6 C/m^2