CAPA CITORS WITH TWO DIELECTRICS GOAL Derive a symbolic expression for a parallel-plate capacitor with two dielectrics. PROBLEM A parallel-plate capacitor has dielectrics with constants κ1 and κ2 between the two plates, as shown in Figure 16.29. Each dielectric fills exactly half the volume between

Question

CAPA CITORS WITH TWO DIELECTRICS

GOAL Derive a symbolic expression for a parallel-plate capacitor with two dielectrics.

PROBLEM A parallel-plate capacitor has dielectrics with constants [latex]\kappa_{1}[/latex] and [latex]\kappa_{2}[/latex] between the two plates, as shown in Figure 16.29. Each dielectric fills exactly half the volume between the plates. Derive expressions for (a) the potential difference between the two plates and (b) the resulting capacitance of the system.

STRATEGY The magnitude of the potential difference between the two plates of a capacitor is equal to the electric field multiplied by the plate separation. The elec tric field in a region is reduced by a factor of [latex]1 / \kappa[/latex] when a dielectric is introduced, so [latex]E=\sigma / \epsilon=\sigma / \kappa \epsilon_{0}[/latex]. Add the potential difference across each dielectric to find the total potential difference [latex]\Delta V[/latex] between the plates. The voltage difference across each dielectric is given by [latex]\Delta V=E d[/latex], where [latex]E[/latex] is the electric field and [latex]d[/latex] the displacement. Obtain the capacitance from the relationship [latex]C=Q / \Delta V[/latex].

Accepted Answer

(a) Derive an expression for the potential difference between the two plates.

Write a general expression for the potential difference across both slabs:

[latex]\Delta V=\Delta V_{1}+\Delta V_{2}=E_{1} d_{1}+E_{2} d_{2}[/latex]

Substitute expressions for the electric fields and dielectric thicknesses, [latex]d_{1}=d_{2}=d / 2[/latex] :

[latex]\Delta V=\frac{\sigma}{\kappa_{1} \epsilon_{0}} \frac{d}{2}+\frac{\sigma}{\kappa_{2} \epsilon_{0}} \frac{d}{2}=\frac{\sigma d}{2 \epsilon_{0}}\left(\frac{1}{\kappa_{1}}+\frac{1}{\kappa_{2}}\right)[/latex]

(b) Derive an expression for the resulting capacitance of the system.

Write the general expression for capacitance:

[latex]C=\frac{Q}{\Delta V}[/latex]

Substitute [latex]Q=\sigma A[/latex] and the expression for the potential difference from part (a):

[latex]C=\frac{\sigma A}{\frac{\sigma d}{2\epsilon_{0}}\left(\frac{1}{\kappa_{1}}+\frac{1}{\kappa_{2}}\right)}=\frac{2 \epsilon_{0} A}{d} \frac{ \kappa_{1} \kappa_{2}} {\kappa_{1}+\kappa_{2}}[/latex]

REMARKS The answer is the same as if there had been two capacitors in series with the respective dielectrics. When a capacitor consists of two dielectrics as shown in Figure 16.30, however, it's equivalent to two different capacitors in parallel.

Question 16.12: CAPA CITORS WITH TWO DIELECTRICS GOAL Derive a symbolic expr...

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