Questions
A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in
a
Reduction of charge on the plates and increase of potential difference across the plates
b
Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates
c
Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates
d
None of the above
detailed solution
Correct option is C
Battery in disconnected so Q will be constant. As C ∝K, so with introduction of dielectric slab capacitance will increase. Using Q = CV, V will decrease and using U=Q22C, energy will decreaseTalk to our academic expert!
Similar Questions
What is the magnitude of charge that will reside in the parallel plate capacitor formed by these two plates.
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
Popular Books
HC Verma Part 1HC Verma Part 2RS AgarwalRD SharmaLakhmir SinghDK GoelSandeep GargTS GrewalNCERT solutions
Class 12 NCERT SolutionsClass 11 NCERT SolutionsClass 10 NCERT SolutionsClass 9 NCERT SolutionsClass 8 NCERT SolutionsClass 7 NCERT SolutionsClass 6 NCERT Solutions
