Problem 13.127 Challenge

จาก Gauss’s Law จะได้สนามไฟฟ้าระหว่างทรงกลมทั้งสองบนแนวแกน x เป็น

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

îî

\(\vec{E} =\frac{Q}{4\pi ε_{0}x^{2}}+\frac{Q}{4\pi ε_{0}(d - x)^{2}}\)

î

\(\vec{E} =\frac{Q}{4\pi ε_{0}}(\frac{1}{x^{2}}+\frac{1}{(d - x)^{2}})\)

\(_{+}\)

จาก\(|ΔV|=-\int \vec{E}·d\vec{r}\)

\(_{-}\)

\(^{+}îîĵk̂\)

\(|\Delta V|=-\int (\frac{Q}{4\pi ε_{0}}(\frac{1}{x^{2}}+\frac{1}{(d - x)^{2}}) )·(dx + dy + dz )\)

\(_{-}\)

\(_{a}\)

\(|\Delta V| = -\int \frac{Q}{4\pi ε_{0}}(\frac{1}{x^{2}}+\frac{1}{(d - x)^{2}})dx\)

\(_{d-b}\)

\(_{a}\)

\(|\Delta V| =\frac{Q}{4\pi ε_{0}}(\frac{1}{x}-\frac{1}{d - x})\)

\(_{d-b}\)

\(|\Delta V| =\frac{Q}{4\pi ε_{0}}(\frac{1}{a}-\frac{1}{d - a}-\frac{1}{d - b}+\frac{1}{d - (d - b)})\)

\(|\Delta V| =\frac{Q}{4\pi ε_{0}}(\frac{1}{a}+\frac{1}{b}-\frac{1}{d - a}-\frac{1}{d - b})\)

จาก d ≫ a จะได้ d - a ≈ d

และ d ≫ b จะได้ d - b ≈ d

\(\therefore |\Delta V| =\frac{Q}{4\pi ε_{0}}(\frac{1}{a}+\frac{1}{b}-\frac{1}{d}-\frac{1}{d})\)

\(|\Delta V| =\frac{Q}{4\pi ε_{0}}(\frac{1}{a}+\frac{1}{b}-\frac{2}{d})\)

จาก \(C = \frac{|Q|}{|\Delta V|}\)

\(C =\frac{Q}{\frac{Q}{4\pi ε_{0}}(1+1-2)}\)

abd

\(C =\frac{4\pi ε_{0}}{\frac{1}{a}+1-2}\;\blacksquare\)

bd