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

$A$ parallel plate capacitor has circular plates of $0.08\,m$ radius and $1.0 \times 10^{-3}\,m$ separation. If a potential difference of $100\,V$ is applied,the charge on the capacitor will be:

$A$ capacitor is made of a flat plate of area $A$ and a second plate having a stair-like structure as shown in the figure. If the area of each stair is $\frac{A}{3}$ and the height of each step is $d$,the capacitance of the arrangement is:

How will the voltage $(V)$ between the two plates of a parallel plate capacitor depend on the distance $(d)$ between the plates,if the charge on the capacitor remains the same?

$A$ capacitor is made of two square plates each of side $a$ making a very small angle $\alpha$ between them,as shown in figure. The capacitance will be close to

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?

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