Physics, asked by Austinnialayatt1491, 9 months ago

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

Answered by princetyagi368
1

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

Answered by qwwestham
5

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

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