Question 23.4: The parallel plates in Fig. 23.9a have an area A = 0.2 m² an......
The parallel plates in Fig. 23.9a have an area A = 0.2 m² and separation distance d = 0.01 m. The original potential difference between them is ΔV_{o} = 300 V which decreases to ΔV = 100 V when a dielectric sheet fills the space between the plates, see Fig. 23.9b. (a) Calculate the capacitance C_{o} , the magnitude of the charge Q_{o} , and the magnitude of the electric field E_{o} . (b) Calculate the final capacitance C and the dielectric constant κ. (c) Find the magnitudes of the induced charge density σ_{i}, the induced electric field E_{i}, and the final electric field E.
(a) Using the parallel-plate capacitor Eq. 23.5, we get:
C={\frac{\epsilon_{o}A}{d}} (Parallel-plate capacitor) (23.5)
C_{o}={\frac{\epsilon_{o}A}{d}}={\frac{(8.85\times10^{-12}\,\mathrm{F/m})(0.2\,\mathrm{m}^{2})}{0.01\,\mathrm{m}}}=1.77\times10^{-10}\,\mathrm{F}=177\,\mathrm{pF}Then, when using Eq. 23.1, the magnitude of the charge on each plate will be the following:
C={\frac{Q}{\Delta V}} (23.1)
Q_{\circ}=C_{\circ}\,\Delta V_{\circ}=(1.77\times10^{-10}\,\mathrm{F})(300\,\mathrm{V})=5.31\times10^{-8}\,\mathrm{C}=53.1\,\mathrm{nC}Finally, we use Eq. 22.17 to find the magnitude of the uniform electric field E_{\circ} as follows:
ΔV = V_{B} − V_{A} = Ed (Opposite the field) (22.17)
E_{o}={\frac{\Delta V_{o}}{d}}={\frac{300\,\mathrm{V}}{0.01\,\mathrm{m}}}=3\times10^{4}\,\mathrm{V/m}Alternatively, we can use the relation E_{\circ} = σ_{\circ}/\epsilon_{\circ} to find E_{\circ}. First, we calculate σ_{\circ} as follows:
\sigma_{o}={\frac{Q_{o}}{A}}={\frac{5.31\times10^{-8}\,\mathrm{C}}{0.2\,\mathrm{m}^{2}}}=2.655\times10^{-7}\,\mathrm{C/m}^{2}Then we find the value of E_{\circ}as follows:
E_{\circ}={\frac{\sigma_{\circ}}{\epsilon_{\circ}}}={\frac{2.655\times10^{-7}\,\mathrm{C/m}^{2}}{8.85\times10^{-12}\,\mathrm{F/m}}}=3\times10^{4}\,\mathrm{C/F}\mathrm{m}=3\times10^{4}\,\mathrm{V/m}(b) We first use Eq. 23.1 to find C as follows:
C={\frac{Q_{\circ}}{\Delta V}}={\frac{5.31\times10^{-8}\mathrm{{\scriptsize~C}}}{100\mathrm{{\scriptsize~V}}}}=5.31\times10^{-10}\,{\mathrm{F}}=531\,{\mathrm{pF}}Then, by using equation C = κC_{\circ}, we find that:
\kappa={\frac{C}{C_{\circ}}}={\frac{5.31\times10^{-10}\,{\mathrm{F}}}{1.77\times10^{-10}\,{\mathrm{F}}}}=3(c) The induced charge density σ_{i} can be obtained from Eq. 23.20 as follows:
{{{\sigma_{i}={\frac{\kappa-1}{\kappa}\sigma_{\circ}}}}} (Parallel-plate capacitor) (23.20)
={\frac{(3-1)(2.655\times10^{-7}\,{\mathrm{C}}/\mathrm{m}^{2})}{(3)}}\;=1.77\times10^{-7}\,{\mathrm{C}}/{\mathrm{m}}^{2}The magnitude of the induced electric field is therefore:
E_{\mathrm{i}}\,={\frac{\sigma_{\mathrm{i}}}{\epsilon_{o}}}={\frac{1.77\times10^{-7}\,\mathrm{C}}{8.85\times10^{-12}\,\mathrm{F/m}}}=2\times10^{4}\,\mathrm{V/m}The magnitude of the final electric field 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}}{3}}=10^{4}\,\mathrm{V/m}Alternatively, we can find E from Eq. 23.18 as follows:
E = E_{\circ} − E_{i} (23.18)
=3\times10^{4}\,\mathrm{V/m}-2\times10^{4}\,\mathrm{V/m}=\mathrm{~10^{4}~V/m}