Question 5.9: In a parallel plate capacitor with two dielectric slabs, the......
In a parallel plate capacitor with two dielectric slabs, the lengths of the dielectric slabs are 3 mm and 5 mm, respectively, while their associated permittivities are \varepsilon _{1} = 2 and \varepsilon _{2} = 4, respectively. A voltage of 100 V is applied across the parallel plates, where the area of each plate is 0.56× 10^{-6} m^{2}. Determine the (i) capacitance, (ii) q, (iii) D, (iv) E _{1}, E _{2} and (v) V _{1}, V _{2}.
(i) The capacitance can be determined as,
C=\frac{\varepsilon _{0}S}{\frac{l_{1}}{\varepsilon_{1}}+\frac{l_{2}}{\varepsilon _{2}} }=\frac{8.854\times 10^{-12}\times 0.56\times 10^{-6}}{\frac{0.003}{2} +\frac{0.005}{4}}=1803\times 10^{-18} F (5.133)
(ii) The charge is,
q=C V=0.0018\times100=0.18\ \mathrm{pC} (5.134)
(iii) The electric flux density can be determined as,
D={\frac{q}{S}}={\frac{0.18\times10^{-12}}{0.56\times10^{-6}}}=0.321\times10^{-6}~\mathrm{C/m^{2}} (5.135)
(iv) The electric field intensities are,
E_{1}={\frac{D}{\varepsilon_{0}\varepsilon _{1}}}={\frac{0.321\times10^{-6}}{2\times8.854\times10^{-12}}}=18127.4{\mathrm{~V/m}} (5.136)
E_{2}={\frac{D}{\varepsilon_{0}\varepsilon_{2}}}={\frac{0.321\times10^{-6}}{4\times8.854\times10^{-12}}}=9063.7\mathrm{~V/m} (5.137)
(v) The voltages can be determined as,
V_{1}=E_{1}l_{1}=18127.4\times003=54.38\ \mathrm{V} (5.138)
V_{2}=E_{2}l_{2}=9063.7\times0.005=45.32\ \mathrm{V} (5.139)