In an insulated parallel-plate capacitor of capacitance $C$,the four surfaces have charges $Q_1, Q_2, Q_3$ and $Q_4$ as shown. The potential difference between the plates is

  • A
    $\frac{Q_2 - Q_3}{2C}$
  • B
    $\frac{Q_2 + 2Q_3}{2C}$
  • C
    $\frac{Q_1 + Q_4}{2C}$
  • D
    $\frac{Q_1 + Q_2 + Q_3 + Q_4}{2C}$

Explore More

Similar Questions

The capacity of a parallel plate capacitor increases with the

The area of each plate of a parallel plate capacitor is $20 \, cm^2$ and the separation between the plates is $2 \, mm$. If the dielectric strength of air is $3 \times 10^6 \, V/m$,the maximum possible value of the emf of the battery that can be connected across the plates of this capacitor and the corresponding charge on the plates are:

What is the area of the plates of a capacitor with a capacitance of $3 \ F$? The distance between the two plates is $5 \ mm$.

Five conducting parallel plates having area $A$ and separation between them $d$ are placed as shown in the figure. Plate number $2$ and $4$ are connected by a wire,and between point $A$ and $B$,a cell of emf $E$ is connected. The charge flown through the cell is

Difficult
View Solution

Assertion : The total charge stored in a capacitor is zero.
Reason : The field just outside the capacitor is $\frac{\sigma }{{{\varepsilon _0}}}$. ( $\sigma $ is the charge density).

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