Problem 13.128 Challenge

จาก Gauss’s Law จะได้สนามไฟฟ้าระหว่างทรงกลมทั้งสองบนแนวแกน x เป็น

\(\vec{E} = \vec{E}_{+}+ \vec{E}_{-}\)

îî

\(\vec{E} =\frac{Q}{2\pi ε_{0}xL}+\frac{Q}{2\pi ε_{0}(d - x)L}\)

î

\(\vec{E} =\frac{Q}{2\pi ε_{0}L}(\frac{1}{x}+\frac{1}{d - x})\)

\(_{+}\)

จาก\(|ΔV|=-\int \vec{E}·d\vec{r}\)

\(_{-}\)

\(^{+}îîĵk̂\)

\(|\Delta V|=-\int (\frac{Q}{2\pi ε_{0}L}(\frac{1}{x}+\frac{1}{d - x}) )·(dx + dy + dz )\)

\(_{-}\)

\(_{a}\)

\(|\Delta V| = -\int \frac{Q}{2\pi ε_{0}L}(\frac{1}{x}+\frac{1}{d - x})dx\)

\(_{d-b}\)

\(_{a}\)

\(|\Delta V| =\frac{Q}{2\pi ε_{0}L}(\ln (d-x) - \ln (x))\)

\(_{d-b}\)

\(_{a}\)

\(|\Delta V| =\frac{Q}{2\pi ε_{0}L}\ln (\frac{d - x}{x})\)

\(_{d-b}\)

\(|\Delta V| =\frac{Q}{2\pi ε_{0}L}(\ln (\frac{d - a}{a}) - \ln (\frac{d - (d - b)}{d - b}))\)

\(|\Delta V| =\frac{Q}{2\pi ε_{0}L}(\ln (\frac{d - a}{a}) - \ln (\frac{b}{d - b}))\)

\(|\Delta V| =\frac{Q}{2\pi ε_{0}L}\ln ((\frac{d - a}{a})(\frac{d - b}{b}))\)

\(|\Delta V| =\frac{Q}{2\pi ε_{0}L}\ln (\frac{(d - a)(d - b)}{ab})\)

จาก d ≫ a จะได้ d - a ≈ d

และ d ≫ b จะได้ d - b ≈ d

\(\therefore |\Delta V| =\frac{Q}{2\pi ε_{0}L}\ln (\frac{(d)(d)}{ab})\)

\(|\Delta V| =\frac{Q}{2\pi ε_{0}L}\ln (\frac{d^{2}}{ab})\)

จาก \(C = \frac{|Q|}{|\Delta V|}\)

\(C =\frac{Q}{\frac{Q}{2\pi ε_{0}L}\ln (d^{2})}\)

ab

\(C =\frac{2\pi ε_{0}L}{\ln (\frac{d^{2}}{ab})}\;\blacksquare\)