1
\$\begingroup\$

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:

enter image description here

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

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

(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?

\$\endgroup\$
3
  • \$\begingroup\$ Have you taken into account that inserting the dielectric requires force and, that force multiplied by the distance is work and that requires energy to be taken from the electric field? \$\endgroup\$
    – Andy aka
    Feb 28 at 11:49
  • \$\begingroup\$ You get the same answer with K=1. But different when K ne 1. You are taking two snapshots from two different methods. But there is external energy involved in inserting a dielectric -- not accounted for in your analysis. It's not a closed system comparison. You should ask this on the physics site, though. Here we treat things in bulk, as you are. But there they are more nuanced. Note also that there field lines that also go between the plates outside of the dielectric. It's not all between the plates in simple lines. There is the rest of the story in the lines going through the air/vacuum. \$\endgroup\$ Feb 28 at 11:49
  • \$\begingroup\$ The charge accumulated in the section of width x is different from that in the section L-x due to the presence of the different dielectric. Instead you used equal charge on both surfaces. \$\endgroup\$
    – Franc
    Feb 28 at 13:32

1 Answer 1

0
\$\begingroup\$

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 :

enter image description here

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 :

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

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

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

\$\endgroup\$
1
  • \$\begingroup\$ If the plates are metallic, then the charge density on them is constant. It is not physically possible for there to be two regions with different charge densities. This can only happen if the two plates are not metallic but insulating for example silicon dioxide, glass or plastic. \$\endgroup\$
    – Franc
    Feb 29 at 6:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.