Two parallel plates have equal and opposite charge. When the space between them is evacuated, the electric field between the plates is 2 × 105 Vm - 1. When the space is filled with dielectric thel electric field becomes 1 × 105 Vm -1 . The dielectric constant of dielectric material is

Question

Two parallel plates have equal and opposite charge. When the space between them is evacuated, the electric field between the plates is 2 × 105 Vm - 1. When the space is filled with dielectric thel electric field becomes 1 × 105 Vm -1 . The dielectric constant of dielectric material is (A) 1/2 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

K = ( E without dielectric / E with dielectric) = ( 2 × 105 / 1 × 105 ) = 2

Q. Two parallel plates have equal and opposite charge. When the space between them is evacuated, the electric field between the plates is 2×105Vm−1. When the space is filled with dielectric thel electric field becomes 1×105Vm−1 . The dielectric constant of dielectric material is

1/2

1

2

3

K = Ewithdielectric​Ewithoutdielectric​​=1×1052×105​ = 2

K = $\frac{E _{w i t h o u t d i e l ec t r i c}}{E _{w i t h d i e l ec t r i c}} = \frac{2 \times 1 0 ^{5}}{1 \times 1 0 ^{5}}$ = 2