Question

what is the dielectric constant of a dielectric on which the induced bound charge density is 69% of the free charge density on the plates of a capacitor filled by the dielectric?

Answer 1

Step-by-step explanation:

The relationship between the bound charge density and the free charge density in a dielectric material is given by:

ρb = ρf (εr - 1)

where:

ρb is the bound charge density

ρf is the free charge density

εr is the relative permittivity or dielectric constant of the material

In this case, we know that the induced bound charge density is 69% of the free charge density, or:

ρb = 0.69 ρf

Plugging this into the equation above, we get:

0.69 ρf = ρf (εr - 1)

Simplifying, we get:

εr = 1 + 0.69

εr = 1.69

Therefore, the dielectric constant of the dielectric is 1.69.