The left plate of the capacitor shown in the figure carries a charge +Q while the right plate is uncharged. The total final charge on the right plate after closing the switch S will be
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
Q2+Cε
b
Q2-Cε
c
-Q2
d
none of these
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Electric field between the capacitor platesE=Q+x2Aε0+x2Aε0=12ε0AQ+2xPotential difference =Ed=d2ε0AQ+2x=εε=Q+2x2C ⇒x=Q2-Cε
Not sure what to do in the future? Don’t worry! We have a FREE career guidance session just for you!
The left plate of the capacitor shown in the figure carries a charge +Q while the right plate is uncharged. The total final charge on the right plate after closing the switch S will be