Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
106 views
in Physics by (15 points)
The capacitance of a parallel plate capacitor is c0 when the plates has air between them. This region is now filled with dielectric slab of dielectric constant K capacitor is connected with battery of EMF and E and Zero internal resistance. Now slab is taken out, then during removal of the slab

Please log in or register to answer this question.

1 Answer

0 votes
by (24.8k points)
  •  Charge CE(k−1) flows through the cell 
  • The external agent has to do 2 1 ​ E 2 C(k−1) amount of work to take the slab out 

After inserting the dielectric the capacitance will be C ′ =kC 

The charge flow through the cell when the slab insert is Q ′ =C ′ E=kCE 

The charge flow through the cell when the slab is taken out =Q=CE 

Net charge flow through the cell Q n ​ =Q ′ −Q=kCE−CE=CE(k−1) 

The amount of work for taking out slab is W= 2 1 ​ (CkE2 −CE2)=1/2CE2(k−1)

No related questions found

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...