Potential energy of a uniformly charged parallel plate capacitor filled with dielectric

Question

For a uniformly charged parallel plate capacitor (dimensions l × w × d), its potential energy is given by:

$$ U = \frac{Q^2}{2C}\ $$

Now if we insert a material (dimension x × w × d) of dielectric constant K in absence of a battery, then its potential energy is:

$$C_{eq} = \frac{(L-x)\ w\ \varepsilon_0}{d}\ + \frac{K\ x \ w \ \varepsilon_0}{d}\ $$

$$C_{eq} = \frac{(L-x+xK) \ w\ \varepsilon_0}{d}\ $$

(where Ceq is equivalent capacitance of system)

$$ U = \frac{Q^2}{2C_{eq}}\ $$

$$ U = \frac{d\ Q^2}{2(L-x+xK)\ w\ \varepsilon_0}\ .......... (i)$$

But if I derive an expression for potential energy using energy density, it comes out different:

$$ U = \frac{1}{2}\varepsilon_0E^2V\ $$

(let's assume potential energy stored in part of capacitor filled with air (or vacuum) = U1 and with dielectric = U2)

$$ U = U_1 + U_2 $$ $$ U = \frac{1}{2}\varepsilon_0\Big(\frac{σ}{\varepsilon_0}\Big)^2(L-x)\ w\ d\ + \frac{1}{2}K\ \varepsilon_0\Big(\frac{σ}{K\varepsilon_0}\Big)^2\ x\ w\ d\ $$

$$ U = \frac{1}{2}\varepsilon_0\Big(\frac{σ}{\epsilon_0}\Big)^2\Big(L-x+\frac{x}{K}\Big)\ w\ d\ $$ where σ is surface charged density of plates; \$σ = Q/(L \cdot w)\$

$$ U = \frac{1}{2\varepsilon_0}\Big(\frac{Q}{L\cdot w}\Big)^2\Big(L-x+\frac{x}{K}\Big)\ w\ d\ .......... (ii)$$

Why are both expressions different?

Answer 1

Thank you so much @Franc for highlighting where I was wrong. Actually I assumed charge density is uniform even after inserting dielectric but to maintain same potential difference across the plate charge density will be different. Here's rest of solution :

Let final Potential difference is V and surface charge densities are σ₁ and σ₂.

$$ V = \frac{σ_2}{\epsilon_o}d\ = \frac{σ_1}{K\epsilon_o}d\ $$

$$ K{σ_2} = {σ_1} $$

Charge Conservation :

$$ {σ_1}xw + {σ_2}(l-x)w = Q $$ $$ K{σ_2}xw + {σ_2}(l-x)w = Q $$ $$ {σ_2} = \frac{Q}{w(l-x+xk)}\ $$

Electric field :

$$ E = \frac{σ_2}{\epsilon_o}\ $$ $$ E = \frac{Q}{w(l-x+xk)\epsilon_o}\ $$

Potential Energy :

$$ U = U_1 + U_2 $$ $$ U = \frac{1}{2}\epsilon_0E^2(l-x)wd\ + \frac{1}{2}K\epsilon_0E^2xwd\ $$

$$ U = \frac{1}{2}\epsilon_0E^2(l-x+xK)wd\ $$

$$ U = \frac{dQ^2}{2(l-x+xK)w\epsilon_0}\ $$