Charges of 2Q and -Q are placed on two plates | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For a parallel plate capacitor with charges $Q_1$ and $Q_2$ on the two plates,the charge on the inner surfaces is given by $q = \frac{Q_1 - Q_2}{2

Vedclass · Accepted Answer

$V = \frac{3Q}{2C}$

Vedclass · Answer

(B) For a parallel plate capacitor with charges $Q_1$ and $Q_2$ on the two plates,the charge on the inner surfaces is given by $q = \frac{Q_1 - Q_2}{2}$.Here,$Q_1 = 2Q$ and $Q_2 = -Q$.Therefore,the charge on the inner surface of the first plate is $q = \frac{2Q - (-Q)}{2} = \frac{3Q}{2}$.The charge on the inner surface of the second plate is $-q = -\frac{3Q}{2}$.The potential difference $V$ between the plates is given by the relation $q = CV$,where $q$ is the magnitude of the charge on the inner surface.Substituting the values,we get $\frac{3Q}{2} = CV$.Thus,$V = \frac{3Q}{2C}$.

Charges of $2Q$ and $-Q$ are placed on two plates of a parallel plate capacitor. If the capacitance of the capacitor is $C$,find the potential difference between the plates.

Similar Questions

The electrostatic force between the metal plates of an isolated parallel plate capacitor $C$ having a charge $Q$ and area $A$ is

The capacity of a parallel plate capacitor is $15\,\mu F$,when the distance between its plates is $6\,cm$. If the distance between the plates is reduced to $2\,cm$,then the capacity of this parallel plate capacitor will be......$\mu F$.

If the charge on parallel plates is $\pm Q$ and the area is $A$,write the formula for the electric field between the two plates.

$A$ parallel plate capacitor is made of two plates with a diameter of $4 \ cm$. What should be the distance between the two plates so that its capacitance is equal to the capacitance of a sphere with a diameter of $20 \ cm$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module