a) less than ρ/ε₀ N/C
b) E=ρ/ε₀ N/C
c) E>ρ/2ε₀ N/C
d) E=ρ/2ε₀ N/C
Doubt by Ananya
Solution :
We know, electric field between two equal and oppositely charged plates having vacuum between them will be given by
E₀=ρ/ε₀ — (1)
But when the space between them is filled with a dielectric substance mica then
E=ρ/Kε₀
where K is the dielectric constant of mica.
E=ρ/Kε₀ OR
E = ρ/ε — (2)
Dividing (2) by (1)
E = ρ/ε — (2)
Dividing (2) by (1)
E/E₀=(ρ/ε)/ρ/ε₀
E/E₀=ε₀/ε
E/E₀=ε₀/ε
E/E₀=1/K [∵K=ε/ε₀]
E=E₀/K
Clearly
E<E₀
E<ρ/ε₀ N/C
E=E₀/K
Clearly
E<E₀
E<ρ/ε₀ N/C
Hence, the new electric field in the region between the plates will be less than ρ/ε₀ N/C.
Hence, a) less than ρ/ε₀ N/C will be the correct option.