Q. 325.0( 2 Votes )

A charge of 1 μC is given to one plate of a parallel-plate capacitor of capacitance 0.1 μF and a charge of 2 μC is given to the other plate. Find the potential difference developed between the plates.

Answer :

Explanation:


Given,


Charge on plate 1, Q1 = 1 μC


Charge on plate 2, Q2 = 2 μC


Capacitance of the capacitor, C= 0.1 μF


Formula used


We know that for a capacitor with net charge, Q and capacitance, C, the Potential difference deceloped in between the plates, V is,



The charges on the inner plates of the capacitor with plates having charges Q1 and Q2 is,



So, the charge, Q by substituting the given values, is



Hence from eqn.1, the potential difference



Hence the potential difference developed in between the plates is 5V.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Interactive Quiz on Dielectric in CapacitorsInteractive Quiz on Dielectric in CapacitorsInteractive Quiz on Dielectric in Capacitors37 mins
Combination of capacitorsCombination of capacitorsCombination of capacitors46 mins
Interactive Quiz on Combination of capacitorsInteractive Quiz on Combination of capacitorsInteractive Quiz on Combination of capacitors49 mins
How does combination of capacitors work?How does combination of capacitors work?How does combination of capacitors work?34 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :