Problem 13.144 Challenge
หัวข้อ: ค่าคงที่ไดอิเล็กทริก (Capacitors with Dielectrics)
\(C =\frac{κ_{1}κ_{2}}{(κ_{2} - κ_{1})}\ln (\frac{κ_{2}}{κ_{1}})\frac{ε_{0}WL}{t}\;\blacksquare ; κ_{new}=\frac{κ_{1}κ_{2}}{(κ_{2} - κ_{1})}\ln (\frac{κ_{2}}{κ_{1}})\)
ดูวิธีทำ
พิจารณาทีละส่วนเล็กๆ กว้าง dx เท่าๆ กัน
จะได้ \(dC_{1} = \frac{κ_{1}ε_{0}Wdx}{\frac{tx}{L}}=\frac{κ_{1}ε_{0}WLdx}{tx}\)
และ \(dC_{2} = \frac{κ_{2}ε_{0}Wdx}{t(1 -\frac{x}{L})}=\frac{κ_{2}ε_{0}WLdx}{t(L - x)}\)
พิจารณาการต่ออนุกรมกันของส่วนเล็กๆ dC₁ กับ dC₂
จะได้ \(\frac{1}{dC}=\frac{1}{dC_{1}}+\frac{1}{dC_{2}}\)
\(dC =\frac{dC_{1}dC_{2}}{dC_{1} + dC_{2}}\)
\(dC =\frac{(\frac{κ_{1}ε_{0}WLdx}{tx})(\frac{κ_{2}ε_{0}WLdx}{t(L - x)})}{(\frac{κ_{1}ε_{0}WLdx}{tx}) + (\frac{κ_{2}ε_{0}WLdx}{t(L - x)})}\)
\(dC =\frac{κ_{1}κ_{2}ε_{0}WL}{(κ_{1}L + (κ_{2} - κ_{1})x)t}dx\)
พิจารณาการต่อขนานของส่วนเล็กๆ dC จากช่วง \(x = 0\) ถึง \(x = L\)
\(_{L}\)
จะได้\(C=\int \frac{κ_{1}κ_{2}ε_{0}WL}{(κ_{1}L + (κ_{2} - κ_{1})x)t}dx\)
\(_{0}\)
\(_{L}\)
\(C =\frac{κ_{1}κ_{2}ε_{0}WL}{t(κ_{2} - κ_{1})}\ln (κ_{1}L + (κ_{2} - κ_{1})x)\)
\(_{0}\)
\(C =\frac{κ_{1}κ_{2}ε_{0}WL}{t(κ_{2} - κ_{1})}(\ln (κ_{1}L + (κ_{2} - κ_{1})L) - \ln (κ_{1}L))\)
\(C =\frac{κ_{1}κ_{2}ε_{0}WL}{t(κ_{2} - κ_{1})}\ln (\frac{κ_{2}}{κ_{1}})\)
\(C =\frac{κ_{1}κ_{2}}{(κ_{2} - κ_{1})}\ln (\frac{κ_{2}}{κ_{1}})\frac{ε_{0}WL}{t}\;\blacksquare ; κ_{new}=\frac{κ_{1}κ_{2}}{(κ_{2} - κ_{1})}\ln (\frac{κ_{2}}{κ_{1}})\)