Two parallel metal plates having charges +Q a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The electric field between two parallel plates is given by $E = \frac{\sigma}{\epsilon_0 K}$,where $\sigma$ is the surface charge density,$\epsilo

Vedclass · Accepted Answer

Decrease

Vedclass · Answer

(B) The electric field between two parallel plates is given by $E = \frac{\sigma}{\epsilon_0 K}$,where $\sigma$ is the surface charge density,$\epsilon_0$ is the permittivity of free space,and $K$ is the dielectric constant of the medium.When the plates are dipped in kerosene oil,the medium between the plates changes from air (or vacuum) to kerosene oil.The dielectric constant $K$ of kerosene oil is greater than $1$ $(K > 1)$.Since the charge $Q$ on the plates remains constant,the surface charge density $\sigma = \frac{Q}{A}$ also remains constant.Therefore,the new electric field $E'$ becomes $E' = \frac{E}{K}$.Since $K > 1$,the electric field $E'$ will be less than the original electric field $E$. Thus,the electric field decreases.

Two parallel metal plates having charges $+Q$ and $-Q$ face each other at a certain distance between them. If the plates are now dipped in a kerosene oil tank,the electric field between the plates will

Similar Questions

For a ray of light,the critical angle is minimum when it travels from

Chemically,hormones are:

If the sum of any two roots of the equation $x^3+p x^2+q x+r=0$ is zero,then

If $\cos ^{-1}\left(\frac{x}{2}\right)+\cos ^{-1}\left(\frac{y}{3}\right)=\theta$,then $9x^2-12xy\cos \theta+4y^2=$

The ratio of minimum wavelengths of Lyman and Balmer series will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module