Problem 13.45 Challenge

พิจารณาสนามไฟฟ้าเนื่องจากจุดประจุ dq ของครึ่งวงกลม

y

R

ds

\(+q\)

dq

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

\(d\vec{E}\)

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

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

\(^{\pi /2}îĵ\)

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

\(_{-\pi /2}\)

\(_{\pi /2}\)

îĵ

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

\(_{-\pi /2}\)

ĵ

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

ĵ

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

ĵ

\(\vec{E} = -\frac{2kq}{\pi R^{2}}\)

พิจารณาสนามไฟฟ้าจากครึ่งวงกลมทั้งสอง

y

x

ĵĵ

\(\vec{E}_{-}= -\frac{2kq}{\pi R^{2}}\vec{E}_{+}= -\frac{2kq}{\pi R^{2}}\)

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

ĵĵ

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

ĵ

\(\vec{E} = -\frac{4kq}{\pi R^{2}}\;\blacksquare\)