Problem 13.112 Challenge
หัวข้อ: การหาศักย์ไฟฟ้าจากสนามไฟฟ้า (Obtaining the Value of the Electric Potential from the Electric Field)
\(V =\frac{σ}{2ε_{0}}(\sqrt{R^{2} + x^{2}}- x) \;\blacksquare ;\) ตรงกับ \(Problem 4.7\)
ดูวิธีทำ
\(_{x}\)
จาก\(V=-\int \vec{E}·d\vec{r}\)
\(_{\infty}\)
\(^{x}îîĵk̂\)
\(V=-\int (\frac{σ}{2ε_{0}}(1 -\frac{x}{R^{2} + x^{2}}) )·(dx + dy + dz )\)
\(_{\infty}\)
\(_{x}\)
\(V = -\frac{σ}{2ε_{0}}\int (1 -\frac{x}{R^{2} + x^{2}})dx\)
\(_{\infty}\)
\(_{xx}\)
\(V = -\frac{σ}{2ε_{0}}( \int dx-\int \frac{x}{R^{2} + x^{2}}dx)\)
\(_{\infty\infty}\)
กำหนดให้ \(u = R^{2} + x^{2}\)
\(\frac{du}{dx}= 2x\)
\(xdx =\frac{du}{2}\)
และ \(x = \infty \) จะได้ \(u = \infty \)
และ \(x = x\) จะได้ \(u = R^{2} + x^{2}\)
\(_{x}_{R^{2} + x^{2}}\)
จะได้ \(V = -\frac{σ}{2ε_{0}}( \int dx-\frac{1}{2}\int \frac{1}{u}du)\)
\(_{\infty\infty}\)
\(_{x}_{R^{2} + x^{2}}\)
\(V = -\frac{σ}{2ε_{0}}(x-u)\)
\(_{\infty\infty}\)
\(_{x}\)
\(V =\frac{σ}{2ε_{0}}(\sqrt{R^{2} + x^{2}}- x)\)
\(_{\infty}\)
\(V =\frac{σ}{2ε_{0}}(\sqrt{R^{2} + x^{2}}-x)-\lim \frac{σ}{2ε_{0}}(\sqrt{R^{2} + x^{2}}- x)\)
\(_{x\to \infty}\)
\(V =\frac{σ}{2ε_{0}}(\sqrt{R^{2} + x^{2}}- x) - 0\)
\(V =\frac{σ}{2ε_{0}}(\sqrt{R^{2} + x^{2}}- x) \;\blacksquare ;\) ตรงกับ \(Problem 4.7\)