The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2 \,\mu F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is.......$\mu F$.
- A$11.2$
- B$15.6$
- C$19.2$
- D$22.4$
Explore More
Similar Questions
$A$ parallel plate capacitor is charged and the charging battery is then disconnected. If the plates of the capacitor are moved further apart by means of insulating handles,then
$A$ parallel plate capacitor is made of two square plates of side $a$,separated by a distance $d$ $(d \ll a)$. The lower triangular portion is filled with a dielectric of dielectric constant $K$,as shown in the figure. The capacitance of this capacitor is
DifficultJEE MAIN 2019
View Solution$A$ parallel plate capacitor is immersed in an oil of dielectric constant $K = 2$. The electric field between the plates is:
Easy
View SolutionTwo 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
$A$ parallel plate air capacitor has capacity $C$ farad, potential $V$ volt, and energy $E$ joule. When the gap between the plates is completely filled with a dielectric of dielectric constant $K$, what happens to the potential $V$ and energy $E$?
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo