Question

Q1

Answer 1

(a)
Generally, \[\overrightarrow{E} = k_{e}\frac{q}{r^2}\hat{r}\]
\[\overrightarrow{E_{Aq}} = k_{e}\frac{3q}{a^2} \hat{Aq}\]\[\overrightarrow{E_{Bq}} = k_{e}\frac{4q}{2a^2} \hat{Bq} = k_{e}\frac{2q}{a^2} \hat{Bq}\]\[\overrightarrow{E_{Cq}} = k_{e}\frac{3q}{a^2} \hat{Cq}\]
\[\overrightarrow{E_{q}} = \overrightarrow{E_{Aq}} +\overrightarrow{E_{Bq}}+\overrightarrow{E_{Cq}}= k_{e}\frac{(3\sqrt{2}+2) q}{a^2} NC^{-1} \], at the direction of θ=45°

(b)
\[\overrightarrow{F_{q}} = \overrightarrow{E_{q}} q = [ k_{e}\frac{(3\sqrt{2}+2) q}{a^2} ]q = k_{e}\frac{(3\sqrt{2}+2) q^2}{a^2} N\]at the direction of θ=45°

Answer 2

(2)

By symmetry, the vertical component of the E-field at O cancels out

\[λ = \frac{(-7.50\times 10^{-6})}{ (12.0/100)} = -6.25\times 10^{-9}Cm^{-1} \]
Genereally,
\[\overrightarrow{E} = \int k_{e} \frac{dq}{r^2}\hat{r}\]

\[\overrightarrow{E_{x}} = \int k_{e} \frac{dq}{r^2}rsin\theta = \frac{k_{e}}{r}\int sin\theta dq\]
But then, I'm stuck. I don't know how to express sinθ in terms of q or l ...

Answer 3

Given : l = 0.12m , q = -7.50μC
radius of the semicircle = 0.12/π
charge density λ = \(\frac{-7.50 \times 10^{-6}}{0.12}= -6.25\times 10^{-9}Cm^{-1}\)
dq = λdt = λrdθ = \]

Answer 4

Given : l = 0.12m , q = -7.50μC
radius of the semicircle = 0.12/π
charge density λ = \(\frac{-7.50 \times 10^{-6}}{0.12}= -6.25\times 10^{-9}Cm^{-1}\)
dq = λdt = λrdθ
\[\overrightarrow{E_{x}} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}k_{e}\frac{dq}{r^2}\hat{r} =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}k_{e}\frac{dq}{r^2}cos \theta =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}k_{e}\frac{\lambda r d\theta}{r^2}cos\theta\]\[=\frac{k_{e}\lambda}r{}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}cos\theta d\theta=\frac{k_{e}\lambda}{r}sin\theta|_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = 2\frac{k_{e}\lambda}{r}=-2940NC^{-1} \]
\[\overrightarrow{E_{y}} = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}k_{e}\frac{dq}{r^2}\hat{r} =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}k_{e}\frac{dq}{r^2}sin \theta =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}k_{e}\frac{\lambda r d\theta}{r^2}sin\theta\]
\[=\frac{k_{e}\lambda}{r}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}sin\theta d\theta=\frac{k_{e}\lambda}{r}cos\theta |_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \=0NC^{-1}\]

So, \[\overrightarrow{E}= \overrightarrow{E_{x}}= -2940NC^{-1}\]
It looks weird though!