When a dielectric material is introduced between plates of a charged condenser, then electric Field between the plates.

Question

When a dielectric material is introduced between plates of a charged condenser, then electric Field between the plates. (A) remain constant (B) decreases (C) increases (D) first (c) then (b)

Tardigrade Admin · Accepted Answer

The electric Field E between the plates of a condenser is given by E=(q/Kε 0A) propto (1/K) Where dielectric medium K is introduced and when K increases the electric field decreases

Q. When a dielectric material is introduced between plates of a charged condenser, then electric Field between the plates.

remain constant

decreases

increases

first (c) then (b)

The electric Field E between the plates of a condenser is given by $E = \frac{q}{K ε _{0} A} \propto \frac{1}{K}$ Where dielectric medium K is introduced and when K increases the electric field decreases

Q. When a dielectric material is introduced between plates of a charged condenser, then electric Field between the plates.

remain constant

decreases

increases

first (c) then (b)

The electric Field E between the plates of a condenser is given by E=Kε0​Aq​∝K1​ Where dielectric medium K is introduced and when K increases the electric field decreases

The electric Field E between the plates of a condenser is given by $E = \frac{q}{K ε _{0} A} \propto \frac{1}{K}$ Where dielectric medium K is introduced and when K increases the electric field decreases