A parallel plate capacitor has plate of length ' l', width ' w ' and separation of plates is 'd '. It is connected to a battery of emf V . A dielectric slap of the same thickness 'd ' and of dielectric constant k = 4 is being, inserted between the plates of the capacitor. At what length of the slab inside plates, will be energy stored in the capacitor be two times the initial energy stored?

Option 1 -

2l/3

Option 2 -

l/2

Option 3 -

l/4

Option 4 -

l/3

3 Views | Posted a month ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

a month ago

Correct Option - 4

Detailed Solution:

Before inserting slab
C_i = ε? A/d
After inserting dielectric slab
C_i = ε? lw/d
C_f = C? + C?
C_f = (Kε? A? /d) + (ε? A? /d)
C_f = 2C_i ⇒ (Kε? wx/d) + (ε? w (l-x)/d) = 2ε? lw/d
4x + l - x = 2l
x = l/3

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