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Question Number 33435 by rahul 19 last updated on 16/Apr/18
In a region an electric field exist in a given direction and it passes through a circle of radius R normally. The magnitude of electric field is given as : E = E_0 (1− (r/R)). where r is the distance from centre of circle .Find electric flux through plane of circle within it.
$$\boldsymbol{{I}}{n}\:{a}\:{region}\:{an}\:{electric}\:{field}\:{exist}\:{in}\:{a}\:{given} \\ $$$${direction}\:{and}\:{it}\:{passes}\:{through}\:{a}\:{circle} \\ $$$${of}\:{radius}\:{R}\:{normally}.\:{The}\:{magnitude} \\ $$$${of}\:{electric}\:{field}\:{is}\:{given}\:{as}\:: \\ $$$$\boldsymbol{{E}}\:=\:{E}_{\mathrm{0}} \:\left(\mathrm{1}−\:\frac{{r}}{\boldsymbol{{R}}}\right).\:{where}\:{r}\:{is}\:{the}\:{distance} \\ $$$${from}\:{centre}\:{of}\:{circle}\:.{Find}\:{electric}\: \\ $$$${flux}\:{through}\:{plane}\:{of}\:{circle}\:{within}\:{it}. \\ $$
Answered by ajfour last updated on 16/Apr/18
φ=∮E^� .dA^� =∫_0 ^( R) E_0 (1−(r/R))(2πrdr) =2πE_0 ∫_0 ^( R) (r−(r^2 /R))dr =2πE_0 ((R^2 /2)−(R^2 /3)) = 2πE_0 ((R^2 /6)) φ = E_0 (((πR^2 )/3)) .
$$\phi=\oint\bar {{E}}.{d}\bar {{A}} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\:\:{R}} {E}_{\mathrm{0}} \left(\mathrm{1}−\frac{{r}}{{R}}\right)\left(\mathrm{2}\pi{rdr}\right) \\ $$$$\:\:\:=\mathrm{2}\pi{E}_{\mathrm{0}} \int_{\mathrm{0}} ^{\:\:{R}} \left({r}−\frac{{r}^{\mathrm{2}} }{{R}}\right){dr} \\ $$$$\:\:\:=\mathrm{2}\pi{E}_{\mathrm{0}} \left(\frac{{R}^{\mathrm{2}} }{\mathrm{2}}−\frac{{R}^{\mathrm{2}} }{\mathrm{3}}\right)\:\:=\:\mathrm{2}\pi{E}_{\mathrm{0}} \left(\frac{{R}^{\mathrm{2}} }{\mathrm{6}}\right) \\ $$$$\:\:\:\phi\:=\:{E}_{\mathrm{0}} \left(\frac{\pi{R}^{\mathrm{2}} }{\mathrm{3}}\right)\:. \\ $$
Commented by rahul 19 last updated on 16/Apr/18
Sir , u have done by considering elemental ring and then integrating from 0→R. But will it not form a disc? and here circle is given( i mean hollow from inside).
$${Sir}\:,\:{u}\:{have}\:{done}\:{by}\:{considering}\: \\ $$$${elemental}\:{ring}\:{and}\:{then}\:{integrating} \\ $$$${from}\:\mathrm{0}\rightarrow{R}.\:{But}\:{will}\:{it}\:{not}\:{form}\:{a}\:{disc}? \\ $$$${and}\:{here}\:{circle}\:{is}\:{given}\left(\:{i}\:{mean}\:{hollow}\right. \\ $$$$\left.{from}\:{inside}\right). \\ $$$$ \\ $$
Commented by ajfour last updated on 16/Apr/18
by mentioning a circle flux is asked to be calculated through the circular flat plane area whose boundary is this circle. Flux is a surface integral.
$${by}\:{mentioning}\:{a}\:{circle}\:\:{flux}\:{is} \\ $$$${asked}\:{to}\:{be}\:{calculated}\:{through}\:{the} \\ $$$${circular}\:{flat}\:{plane}\:{area}\:{whose} \\ $$$${boundary}\:{is}\:{this}\:{circle}. \\ $$$${Flux}\:{is}\:{a}\:{surface}\:{integral}. \\ $$
Commented by rahul 19 last updated on 16/Apr/18
So here if in place of circle , disc is mentioned answer will remain same, right ?
$${So}\:{here}\:{if}\:{in}\:{place}\:{of}\:{circle}\:,\:{disc}\: \\ $$$${is}\:{mentioned}\:{answer}\:{will}\:{remain}\:{same}, \\ $$$${right}\:? \\ $$
Commented by ajfour last updated on 16/Apr/18
yes!
$${yes}! \\ $$

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