NEET PHYSICS by Pradeep kshetrapal
Current Electricity Lesson no. NEET-PHY-11-01
A Parallel plate cacacitor of capacitance C is connected to a cell of emf V and then disconnected from it. A dielectric slab of dielectric constant K which can just full the air gap os the capacitor is now inserted in it. Which of the following is incorrect..
(a) The potential difference the plates deceases K times. (b) The energy stored in the capacitor decrease K times 2 -1) (c) The change in energy stored is CV (d) The charge on the capacitor is now conserved
When battery is disconnected, the path of charge travel (in and out) is closed. Hence charge become a constant quantity. This makes statement (d) as correct statement.
The dielectric plate insertion always increase the capacitance by K times. Now since charge q is constant and C has increased K times, the voltage-will decrease to That is statement (a) that voltage decrease K times is a KC correct statement.
For calculating effect on energy we consider relation as energy stored-- This relation is taken up since q is a constant in this case ane energy depends only on C. 2C Here C has increase K times hence energy decreases K times which shows that statement (b) is correct
Since statements a,b and d are discussed to be correct, the only incorrect statement remain is (c) . So that is our option for incorrect answer.. As such, we know the relation of decrease in energy to be 2 2 2 2c 2kc 2c Therefore our choice for incorrect answer is (C)
Here we cannot use the relation 1 CV2( -1) because C and V both have changed. With the new values it will beV1) C (-1) which does not match with the given statement. Hence this is not correct statement. Therefore this is our choice for incorrect answer.
State topper in school exams, selected in IIT Kharagpur, M.Sc., 30 yrs in teaching.