Problem 13.46 Challenge

พิจารณาสนามไฟฟ้าเนื่องจากจุดประจุdqของ\(\frac{1}{4}\) ของวงกลม

\(d\vec{E}\)

y

\(\frac{\theta}{R\theta}x\)

dq

ds

îĵ

จะได้ \(d\vec{E} = \frac{kdq}{R^{2}}(\sin \theta + \cos \theta )\)

îĵ

\(d\vec{E} =\frac{k\lambda ds}{R^{2}}(\sin \theta + \cos \theta ); dq = \lambda ds\)

îĵ

\(d\vec{E} =\frac{k\lambda Rd\theta}{R^{2}}(\sin \theta + \cos \theta ); ds = Rd\theta\)

\(^{\pi /4}îĵ\)

\(\vec{E} =\frac{k\lambda}{R}\int (\sin \theta + \cos \theta )d\theta\)

\(_{-\pi /4}\)

\(_{\pi /4}\)

îĵ

\(\vec{E} =\frac{k\lambda}{R}(-\cos \theta + \sin \theta )\)

\(_{-\pi /4}\)

ĵ

\(\vec{E} =\frac{2k\lambda}{R}\)

ĵ

\(\vec{E} =\frac{2k}{R}(\frac{q}{\ell}); \lambda =\frac{q}{\ell}\)

ĵ

\(\vec{E} =\frac{2kq}{R\ell}\)

พิจารณาสนามไฟฟ้าจากครึ่งวงกลมทั้งสอง

\(y|\vec{E}_{2}|=\frac{2kq}{\pi R^{2}}\)

\(|\vec{E}_{1}|=\frac{2kq}{\pi R^{2}}\)

\(|\vec{E}_{3}|=\frac{2 2kq}{\pi R^{2}}\)

\(45°45°\)

x

\(\vec{E} = \vec{E}_{1} + \vec{E}_{2} + \vec{E}_{3}\)

îĵî

\(\vec{E} =\frac{2kq}{\pi R^{2}}(-\cos (45°) + \sin (45°) ) +\frac{2kq}{\pi R^{2}}(\cos (45°)\)

ĵîĵ

\(+ \sin (45°) ) +\frac{2 2kq}{\pi R^{2}}(\cos (45°) + \sin (45°) )\)

îĵ

\(\vec{E} =\frac{2kq}{\pi R^{2}}( + 2 ) \;\blacksquare\)