 # A point charge q is placed in the axis of a ring of radius r at a distance x from the centre. The ring has charge q aswell. Find the tension in the ring.

10 years ago

Dear Anwesan,

The charge density is = Q / (2 a)

Remember, our equation for the electric field is for the magnitude of the electric field. Consider a little piece of charge dq as sketched in the diagram. Because that charge dq is there, there is an electric field dE at point P in the direction shown. The component dEx of that electric field along the direction of the axis perpendicular to the plane of the ring is

dEx = dE cos dEx = dE (x/r)

dEx = [k dq/r2] (x/r)

dEx = [k dq/r3] x

dEx = [k x dq/r3]

dEx = [k x/r3] dq

Notice that, with this geometry, once the radius of the ring a is specified and the position x, that fully specifies r. r and x do not change as we integrate over dq.
[[ Remember, SQRT() means "square root of ()" because that is easier for me to type. ]]

r = SQRT(a2 + x2)

r3 = (a2 + x2)3/2

1/r3 = 1/(a2 + x2)3/2    Remember, x and a are not variables.

Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple  to download the toolbar….

So start the brain storming…. become a leader with Elite Expert League ASKIITIANS

Thanks

Aman Bansal