Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A parallel plate capacitor is charged and then isolated. On increasing the plate separation
Charge Potential Capacitance
a. remains constant remains constant decreases
b. remains constant increases decreases
c. remains constant decreases increases
d. increases increases decreases

Electrostatic Potential and Capacitance

Solution:

Isolated capacitor $\Rightarrow Q$ = constant
separation d increase
$\Rightarrow C =$ decrease
$Q = CV \Rightarrow V =$ increase