The force between the plates of a parallel plate capacitor of capacitance $C$ and distance of separation of the plates $d$ with a potential difference $V$ between the plates is

  • A
    $\frac{C V^2}{2 d}$
  • B
    $\frac{C^2 V^2}{2 d^2}$
  • C
    $\frac{C^2 V^2}{d^2}$
  • D
    $\frac{V^2 d}{C}$

Explore More

Similar Questions

In a parallel plate capacitor with air between the plates,each plate has an area of $6 \times 10^{-3} \, m^{2}$ and the distance between the plates is $3 \, mm$. Calculate the capacitance of the capacitor. If this capacitor is connected to a $100 \, V$ supply,what is the charge on each plate of the capacitor?

The area of each plate of a parallel plate capacitor is $100 \, cm^2$ and the distance between the plates is $1 \, mm$. It is filled with mica of dielectric constant $6$. The radius of the equivalent capacity of the sphere will be.......$m$.

$A$ capacitor of capacitance $C$ with a distance $d$ between its plates is connected to a battery of $V$ volts. What is the force acting between the two plates?

Difficult
View Solution

$A$ parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge,potential,and capacitance respectively are:

What is a parallel plate 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