AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

A dielectric slab fills the lower half. region of parallel plate capacitor as shown in figure. Take plate area as A

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Equivalent capacity of the system is $\frac{ϵ_{0} A}{2 d} (1 + K)$

Net charge on lower & upper. half of-left hand plate are different

Net charge on lower half of .the plate is $\frac{K \in_{0} A}{2 d} \times V$

The net charge of lower hallf of the left hahd plate is $\frac{1}{K}$ times the charge on upper half of plate

answer is A, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

This system can be considered as two capacitors in parallel, with $C_{1} = \frac{ϵ_{0} A}{2 d} & C_{2} = \frac{K ϵ_{0} A}{2 d}$

$C_{1}$ is due to upper half of two plates while $C_{2}$ is due to lower half.

As potential difference two capacitors are same, charge would be different as capacities are different.

Watch 3-min video & get full concept clarity

courses

No courses found

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!