Q.

Two parallel plate capacitors C1 and C2 each having capacitance of 10μF are individually charged by a 100 V D.C. source. Capacitor C1 is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor C2 is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor C1 is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be ____V. (Assuming Dielectric constant = 10)

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

answer is 55.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

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

Take Free Test

Detailed Solution

detailed_solution_thumbnail

In C1Q1=10×10×100=10000μC

(Q=KCV)
In C2Q2=10×100=1000μCQ=C2V
Now C1 & C2 kept in parallel
KC1+KC2V=(10000+1000)10C1+C2V=11000C1+C2V=1100V=110020=55 volt 

Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon