8.5 Molecular model of a dielectric  (Page 3/13)

Electrical field and induced surface charge

Suppose that the distance between the plates of the capacitor in [link] is 2.0 mm and the area of each plate is $4.5\times {10}^{\mathrm{-3}}{\text{m}}^2$ . Determine: (a) the electrical field between the plates before and after the Teflon™ is inserted, and (b) the surface charge induced on the Teflon™ surfaces.

Strategy

In part (a), we know that the voltage across the empty capacitor is $V_0=40\text{V}$ , so to find the electrical fields we use the relation $V=Ed$ and [link] . In part (b), knowing the magnitude of the electrical field, we use the expression for the magnitude of electrical field near a charged plate $E=\sigma \text{/}{\epsilon }_0$ , where $\sigma$ is a uniform surface charge density caused by the surface charge. We use the value of free charge $Q_0=8.0\times {10}^{\mathrm{-10}}\text{C}$ obtained in [link] .

Solution

1. The electrical field $E_0$ between the plates of an empty capacitor is
   
   $E_0=\frac{V_0}{d}=\frac{40\text{V}}{2.0\times {10}^{\mathrm{-3}}\text{m}}=2.0\times {10}^4\text{V/m}.$
   
   The electrical field E with the Teflon™ in place is
   
   $E=\frac{1}{\kappa }{E}_0=\frac{1}{2.1}2.0\times {10}^4\text{V/m}=9.5\times {10}^3\text{V/m}\text{.}$
2. The effective charge on the capacitor is the difference between the free charge $Q_0$ and the induced charge $Q_{\text{i}}$ . The electrical field in the Teflon™ is caused by this effective charge. Thus
   
   $E=\frac{1}{{\epsilon }_0}\sigma =\frac{1}{{\epsilon }_0}\frac{Q_0-Q_{\text{i}}}{A}.$
   
   We invert this equation to obtain $Q_{\text{i}}$ , which yields
   
   $\begin{array}{cc}\hfill Q_{\text{i}}& =Q_0-{\epsilon }_{0}AE\hfill \\ & =8.0\times {10}^{\mathrm{-10}}\text{C}-\left(8.85\times {10}^{\mathrm{-12}}\frac{{\text{C}}^2}{\text{N}·{\text{m}}^2}\right)\left(4.5\times {10}^{\mathrm{-3}}{\text{m}}^2\right)\left(9.5\times {10}^3\frac{\text{V}}{\text{m}}\right)\hfill \\ & =4.2\times {10}^{\mathrm{-10}}\text{C}=0.42\text{nC}\text{.}\hfill \end{array}$

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