JEE Main 31-Aug-2021 Shift 2 Solved Paper

Given, capacity of capacitor, C = 200µF
EMF of battery = Potential difference across capacitor, V = 200 V
Dielectric constant of slab, K = 2
U1=

1

2

CV2=

1

2

×200×(200)2µJ=4J
New capacity with dielectric slab,
C′ = KC = 200 × 2 = 400µF
Since, the battery remains connected.
Then, V′ = V = 200 V
Now, electrostatic energy in capacitor with dielectric
U2=

1

2

C′V′=

1

2

×400×(200)2µJ
= 8 J
Hence, change in the electrostatic energy,
ΔU=U2−U1=8−4=4J
Thus, the correct answer is 4.