$A$ parallel plate capacitor has a potential of $20\,kV$ and a capacitance of $2 \times 10^{-4}\,\mu F$. If the area of the plate is $0.01\,m^2$ and the distance between the plates is $2\,mm$,find the dielectric constant of the medium.

  • A
    $4.52$
  • B
    $3.02$
  • C
    $4.12$
  • D
    $5.10$

Explore More

Similar Questions

Two capacitors,each having capacitance $40\,\mu F$,are connected in series. The space between the plates of one of the capacitors is filled with a dielectric material of dielectric constant $K$ such that the equivalent capacitance of the system becomes $24\,\mu F$. The value of $K$ is:

While a capacitor remains connected to a battery and a dielectric slab is inserted between the plates,then:

$A$ parallel plate capacitor has two layers of dielectrics as shown in the figure. This capacitor is connected across a battery. The ratio of the potential difference across the dielectric layers $(V_1/V_2)$ is:

$A$ voltmeter reads $4 \ V$ when connected to a parallel plate capacitor with air as a dielectric. When a dielectric slab is introduced between the plates for the same configuration,the voltmeter reads $2 \ V$. What is the dielectric constant of the material?

$A$ parallel plate capacitor has plates of area $A$ and a separation distance of $100 \ mm$. It contains two dielectric layers: one with a dielectric constant of $10$ and a thickness of $6 \ mm$,and another with a dielectric constant of $5$ and a thickness of $4 \ mm$. Find the capacitance of the capacitor.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo