## Monday, March 22, 2010

### Irodov Problem 3.19

The electric flux through the circle is given by,

Where E is the electric field and dS is the area vector through which E passes. From problem 3.14 we already know that E is given by,

Consider an infinitesimally thin ring section of radius r and width dr. The area vector of this infinitesimally thin ring section is given by,

Hence the flux is given by,

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SIR, HOW CAN YOU CONSIDER RING WITH SMALL CROSS SECTIONS IF IT WOULD HAVE BEEN DISC YOU WOULD HAVE CONSIDERED BUT IT IS NOT A DISC

ReplyDeleteQuestion is trying to ask how much flux is passing thru circle area ... Not thru any disk/any other body so dont misunderstand...!. Its just asking how many electric lines of force are passing thru circle area.( just like amount of water which passes thru a similar ring when you keep it in front of shower perpendicular to water flow) . Hope i helped !

DeleteAyush ur reply was not understandable to me.

Deletesolid circle=disc

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