Capacitance when dielectric is introduced

[logearav]

Homework Statement

a dielectric of dielectric constant 3 fills three fourth of the space between the plates of a parallel plate capacitor. What percentage of the energy is stored in the dielectric.

Homework Equations

U=CV²/2
= epsilon₀AV²/d.2

The Attempt at a Solution

When dielectric is introduced the above equation changes to epsilon₀AV²/(d-t+t/epsilon_r)
I substituted 3d/4 for t since dielectric fills three fourth of space between the plates of capacitor and 3 for relative permeability.
When simplified i got the answer epsilon₀AV²/d
which is twice the value of U that is 2U
In the book the answer is given as 50% which i don't understand. How this 50% comes? Thanks in advance, revered members

 

[cupid.callin]

how did you take V to be constant?
is it given that capacitor is connected to a battery?

 

[logearav]

no sir. its not given. please guide me to solve this problem

 

[cupid.callin]

is that the exact question given to you?
because i think something else should also be given.
ans also, the language of question is new to me

 

[logearav]

yes sir i have the typed the exact question. no other details are given

 

[cupid.callin]

there is only 1 way i got 50% energy increase due to introduction of dielectric.
first let capacitor is isolated (safe to assume)
let it has plate area A and separation d.
Energy, U = Q^2/2C

find energy when there is no dielectric, in terms of epsilon,A,d not in C (pretty easy :) )

then find net capacitance when dielectric is inserted.

now find energy,
*note: now Q will remain same as capacitor is separated.

you'll see that increase in energy is 1/2 of initial energy ... which is stored due to (or in other words, stored inside) capacitor.

 

[logearav]

sir, can i proceed in this way as given in my attachment? Of course i don't know how 50% comes as given in the book? thanks in advance,sir