Question 649: Electrostats & Capacitance

Question

A thin semi-circular ring of radius $$r$$ has a positive charges $$q$$ distributed uniformly over it. The net field $$\overrightarrow E $$ at the center $$O$$ is

Accepted Answer

Let us consider a differential element $$dl.$$ charge on this element.
$$dq = \left( {{q \over {\pi r}}} ight)dl$$
$$ = {q \over {\pi r}}\left( {rd	heta } ight)\,\,\,\,\,$$ (as $$dl = rd	heta $$)
$$ = \left( {{q \over \pi }} ight)d	heta $$
Electric field at $$O$$ due to $$dq$$ is
$$dE = {1 \over {4\pi { \in _0}}}.{{dq} \over {{r^2}}}$$
$$ = {1 \over {4\pi { \in _0}}}.{q \over {\pi {r^2}}}d	heta $$
The component $$dE\cos 	heta $$ will be counter balanced by another element on left portion.
Hence resultant field at $$O$$ is the resultant of the component $$dE\sin 	heta $$ only.
$$	herefore$$ $$E = \int {dE\sin 	heta  = \int\limits_0^\pi  {{q \over {4{\pi ^2}{r^2}{ \in _0}}}} } \sin 	heta d	heta $$
$$ = {q \over {4{\pi ^2}{r^2}{ \in _0}}}\left[ { - \cos 	heta } ight]_0^\pi $$
$$ = {q \over {4{\pi ^2}{r^2}{ \in _0}}}\left( { + 1 + 1} ight)$$
$$ = {q \over {2{\pi ^2}{r^2}{ \in _0}}}$$
The directions of $$E$$ is towards negative $$y$$-axis.
$$	herefore$$ $$\overrightarrow E  =  - {q \over {2{\pi ^2}{r^2}{ \in _0}}}\widehat j$$

Answer

$${q \over {4{\pi ^2}{\varepsilon _0}{r^2}}}\,j$$

Answer

$$ - {q \over {4{\pi ^2}{\varepsilon _0}{r^2}}}\,j$$

Answer

$$ - {q \over {2{\pi ^2}{\varepsilon _0}{r^2}}}\,j$$

Answer

$$  {q \over {2{\pi ^2}{\varepsilon _0}{r^2}}}\,j$$

Electrostats & Capacitance

A thin semi-circular ring of radius $r has a positive charges q distributed uniformly over it. The net field \overrightarrow E at the center O$ is

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