This one is crazy!

Electricity and Magnetism Level 3

A thin wire ring of radius r r carries a charge q q . The ring is oriented parallel to an infinite conducting plane and is separated by a distance l l from it. Find the surface charge density at a point of the plane symmetrical with respect to the ring.

− q × l 2 π × ( l 2 + r 2 ) 3 / 2 \frac{-q \times l}{2\pi \times (l^2 + r^2)^{3/2}} − q × l 3 π × ( l 2 + r 2 ) 3 / 2 \frac{-q \times l}{3\pi \times (l^2 + r^2)^{3/2}} − q × l 4 π × ( l 2 + r 2 ) 3 / 2 \frac{-q \times l}{4\pi \times (l^2 + r^2)^{3/2}} − q × l 5 π × ( l 2 + r 2 ) 3 / 2 \frac{-q \times l}{5\pi \times (l^2 + r^2)^{3/2}}

Rohan Shrothrium by Rohan Shrothrium , 21, India

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1 solution

Aryan Goyat
Jun 29, 2016

firstly you must know the field at an axis of ring(i personally don't and derive it every time)

now according to image charge method make the image of the ring about the plane (we do this as it makes the plane equipotential) this will now give you field at any pt

net field at the symmetrical pt =2 k q l ( r ( 2 ) + l ( 2 ) ) ( 1.5 ) \frac{kql}{(r^(2)+l^(2))^(1.5)}

equate this to electric field near a conductor ie sigma/eo

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Hey can you just put up your solution......? I posted this question hoping someone would help me out with it :P

Rohan Shrothrium - 4 years, 11 months ago

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