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# we found that a ring of charge with radius a and total charge Q,the potential at point P on the ring axis a distance x from the cinter is

## 2 Answers

10 years ago

electric field at any point on the axis of ring is given by

E=kQx/(a2 +x2 )3/2

and we have E=dv/dx         ( v is potential)

dv = Edx lim from 0 to x

V=kQx/(a2 +x2 )1/2

this is the required expression

10 years ago

for a ring radius is 'a' and charge 'q' is given

The chrfe is accumulated only on the circumference of the ring. So on the axis of the ring at pint 'P' the dist from the charge is r=sqrt(a^2+x^2) where 'x' is the distance of 'P' from the center O.

Let us consider a dq charge on circumference at some point M

dEx=dEcosd         (d is the angle b/w the line OP and PM )

=(k(dq)/r^2)*(x/r)

E=int(dE)

=(kx/r^(3/2))int(dq)

=kxq/((a)+(x))^(3/2)

***

And by the concept that V=-dE/dx we can get the formula for V

O.K All the best

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