# Net bound surface charge density of capacitor

Q. In a parallel plate capacitor, two dielectric slabs of thickness 5 cm each are inserted between the plates and a potential of 100 V is applied across it. The value of the net bound surface charge density at the interface of the two dielectrics is ___.

(Expected ans: $$\frac {- 2000}3ε_0$$)

Electric field in capacitor (without dielectric) $$E = 1000$$ V/m

Polarisation $$= χε_0E = (κ-1)ε_0E$$

Polarised charge on plate 1 = $$ε_0E$$

Polarised charge on plate 2 = $$3ε_0E$$

So the net bound surface charge density should be the difference $$2ε_0E = 2000ε_0$$

I have rewritten my answer as a result of the comment made by @BobD.

In the capacitor as shown in the question the potential of the interface relative to the negative terminal of the voltage supply is $$+100/3\,\rm V$$.

If the capacitor had only air inside it then the potential of that corresponding location would have been $$50\,\rm V$$ as shown in the diagrams below.

With air present the upward electric field over the whole region between the plates of the capacitor is $$E_{\rm a} = 50/d$$.

With the dielectric present the net upward electric field is $$E_{\rm f4} = \frac {100}{3d}$$ for the top dielectric and $$E_{\rm f2} = \frac {200}{3d}$$ for the bottom dielectric.

To produce those changes the bound charges of the top dielectric must have produced a downward electric field $$E_{\rm d4} = \frac{100}{3d}$$ and the bound charges of the bottom dielectric must have produced an upward electric field $$E_{\rm d2} = \frac{100}{3d}$$ noting that $$E_{\rm f4}=E_{\rm a4} + E_{\rm d4}$$ and $$E_{\rm f2}=E_{\rm a2} + E_{\rm d2}$$.

Since $$E = \sigma /\epsilon_0$$ the net bound charge density at the interface is $$-2 \times \frac{50}{3d}= -\frac{2000\,\epsilon_0}{3}.$$

• Comments are not for extended discussion; this conversation has been moved to chat.
– Chris
Commented May 27, 2022 at 15:35

Hint: Assume that the term "net bound surface charge density" refers to $$Q/A$$ at the interface. Then I think the best you can do is determine $$Q/A$$ as a function of the relative permittivities $$k_1$$ and $$k_2$$. To do this, use the fact that

$$C_{1}=\frac{k_{1}\epsilon_{o}A}{d}\tag{2}$$

$$C_{2}=\frac{k_{2}\epsilon_{o}A}{d}\tag{3}$$

Where $$\epsilon_o$$ =permittivity of free space = 8.854 x 10$$^{-12}$$ F/m, $$d$$ = 0.05 m as given, and $$A$$ is the area of the interface ("plate" area).

Hope this helps.

• My answer agrees with theirs up to the point where they introduce P1 and P2, which I’m not familiar with Commented May 24, 2022 at 10:12
• There's some real strange things in their solution. For example, they have two different values of $\sigma_1$. On another line they have $\sigma=-\sigma /4$(??) Commented May 24, 2022 at 11:04