A parallel plate capacitor of plate area A and plate separation d is charged to potential V and then the battery is disconnected. A slab of dielectric constant k is then inserted between the plates of the capacitors so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted) and work done on the system in question in the process of inserting the slab, then state incorrect relation from the following
see full answer
High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET
🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya
a
Q=ε0AVd
b
W=ε0AV22kd
c
E=Vkd
d
W=ε0AV22d1−1k
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
After inserting the dielectric slab New capacitance C'=K.C=Kε0AdNew potential difference V'=VKNew charge Q'=C'V'=ε0AVdNew electric field E'=V'd=VKdWork done (W) = Final energy – Initial energyW=12C′V′2−12CV2=12(KC)VK2−12CV2=12CV21K−1=−12CV21−1K=−ε0AV22d1−1K so |W|=ε0AV22d1−1K.