Join Sarthaks eConnect Today - Largest Online Education Community!
0 votes
12 views
asked ago in Physics by (46.1k points)

A parallel plate capacitor of area A and separation d is charged to potential difference V and removed from the charging source. A dielectric slab of constant K = 2, thickness d and area A/2  is inserted, as shown in the figure. Let σ1 be free charge density at the conductor-dielectric surface and σ2 be the charge density at the conductor-vacuum surface.

(A) The electric field have the same value inside the dielectric as in the free space between the plates.

(B) The ratio σ12 

(C) The new capacitance is 3∈0A/2d

(D) The new potential difference is 2/3 V

1 Answer

+1 vote
answered ago by (14.5k points)
selected ago by
 
Best answer

Correct option (A,B,C,D)

Explanation: 

Potential for each plate remain same over whole area. If potential difference between them is, say V' then V' = Ed i.e. E is also same inside the plates. To keep E same, free charge density is changed i.e. charge redistributes itself. To find new capacitance, two capacitors can be taken as connected in parallel. Then

By Q = CV, as Q remains unchanged V is changed to 2/3 V.

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.

One Thought Forever

“There is a close connection between getting up in the world and getting up in the morning.“
– Anon
~~~*****~~~

Categories

...