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.

Anwesan Pal, 12 years ago

Grade:12

2 Answers

Aman Bansal

592 Points

12 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

dE_x = dE cos

dE_x = dE (x/r)

dE_x = [k dq/r²] (x/r)

dE_x = [k dq/r³] x

dE_x = [k x dq/r³]

dE_x = [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.

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Thanks

Aman Bansal

Askiitian Expert

Anwesan Pal

14 Points

12 years ago

I understood your point but what I wanted to know was the tension in an elementary constituent of the ring due to the coulombic force between the point charge and the ring..........