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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 $2ε_0E = 2000ε_0$

Can someone please help?

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2 Answers 2

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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.

enter image description here

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 $-2 \times \frac{50}{3d}= -\frac{2000\,\epsilon_0}{3}.$

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

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

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