Question

When a dielectric slab of dielectric constant \[K\] is inserted between the plates, what is the electric field \[E'\] due to polarisation of the dielectric?

Options

• \[E'=\frac{E}{K}\]

• \[E'=\frac{K}{E}\]

• \[E'=E+K\]

• \[E'=KE\]

Answer 1

Polarisation of the dielectric reduces the electric field between the plates. The reduced field is \[E'=\frac{E}{K}\].