Question 23.5: Assume that the parallel-plate capacitor of Fig. 23.10a has ......
Assume that the parallel-plate capacitor of Fig. 23.10a has a plate area A = 0.2 m², separation distance d = 10^{−2} m, and original potential difference ΔV_{0} = 300 V. A dielectric slab of thickness a = 5 × 10^{−3} m and dielectric constant κ = 2.5 is inserted between the plates as shown in Fig. 23.10b. (a) Find the magnitudes of the final electric field E in the slab, the final potential difference ΔV between the plates, and the final capacitance C with the dielectric slab in place. (b) Find an expression for C in terms of C_{0}, a, d, and κ.
(a) From Example 23.4, we have E_{0} = 3 × 10^{4} V/m. Therefore, the magnitude of the final electric field in the slab can be obtained from Eq. 23.16 as follows:
\vec{E}=\frac{\vec{E_{\circ}}}{\kappa} (23.16)
E={\frac{E_{\circ}}{\kappa}}={\frac{3\times10^{4}\ {\mathrm{V/m}}}{2.5}}=1.2\times10^{4}\ {\mathrm{V/m}}By applying Eq. 22.6,
\Delta V=V_{B}-V_{A}={\frac{\Delta U}{q}}=-\frac{W_{A B}}{q}=-\int\limits_{A}^{B}{\vec{E}}•d{\vec{s}} (22.6)
we can find ΔV by integrating against the electric field along a straight line from the negative plate (−) to the positive plate (+). Within the dielectric, we must note that \vec{E} • d \vec{s} = −E ds, the path length is a, and the magnitude of the field is E. But within the right and left gaps, the total path length is d − a and the magnitude of the field is E_{\circ}. Thus, Eq. 22.6 yields:
\Delta V=V_{+}-V_{-}=-\int\limits_{-}^{+}\vec{E}• d\vec{s}\,=\,\int\limits_{-}^{+}E\,d s=E_{o}\,(d-a)+E a=(3\times10^{4}\mathrm{~V/m})(10^{-2}\mathrm{~m-5}\times10^{-3}\mathrm{~m})+(1.2\times10^{4}\mathrm{~V/m})(5\times10^{-3}\mathrm{~m})
= 210 V
From Example 23.4, we found that Q_{o} = 5.31 × 10^{−8} C and from Eq. 23.1 we can find the value of C as follows
C={\frac{Q}{\Delta V}} (23.1)
C={\frac{Q_{o}}{\mathrm{{\Delta}}\,V}}={\frac{5.31\times10^{-8}\,\mathrm{{C}}}{\,\mathrm{{210}}\,\mathrm{{V}}}}=2.53\times10^{-10}\,\mathrm{{F}}=0.253\,\mathrm{{nF}}Note that we cannot use the relation C = κ C_{o}, because it is true only if the dielectric material fills the space between the capacitor’s plates.
(b) We start with the proven formula of part (a); that is:
ΔV = E_{0}(d − a) + Ea
Then, using \Delta V=Q_{\circ}/C,\ E_{\circ}=\sigma_{\circ}/\epsilon_{\circ}=Q_{\circ}/\epsilon_{\circ}A,\ C_{\circ}=\epsilon_{\circ}A/d,\,\mathrm{and}\ E=E_{\circ}/\kappa=\sigma_{\circ}/ \epsilon, we can find an expression for C by performing the following steps:
{\frac{1}{C}}={\frac{d-a}{\epsilon_{o}A}}+{\frac{a}{\kappa\epsilon_{o}A}}
C=\frac{\epsilon_{o}A}{\left[(d-a)+\frac{a}{κ}\right]}\quad\Rightarrow\quad C=\frac{d}{\left[(d-a)+\frac{a}{κ}\right]}\frac{\epsilon_{o}A}{d}
C={\frac{d}{\left[(d-a)+{\frac{a}{κ}}\right]}}C_{\circ}
In the second step,(d−a)/\epsilon_{\circ} A is the inverse of the capacitance of an air capacitor of separation d − a, and a/κ\epsilon_{\circ} A is the inverse of the capacitance of a capacitor of separation a but filled with a dielectric.