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There is a fixed semicircular ring of ra...

There is a fixed semicircular ring of radius`R` lying in `yz` plane ,with centre at origin and it is uniformly charged with charge `+Q` A pipe is fixed along `x-axis` from the origin ,The inner surface is pipe is smooth and is made of insultaing material .A small ball of charge `+q` And mass `m` is projected in the pipe with negligible velocity ,ball can move in the pipe whole arrangement is in gravity free space.

Normal reaction exerted by pipe on ball when the ball is moving in pipe is along

A

`+Z` ax is always

B

`Zax` is always

C

initially along y-axis and then along `Z` axis

D

y-axis is always.

Text Solution

Verified by Experts

The correct Answer is:
A


`dF_(x)=q(dE_(x))=q(1)/(4piepsilon_(0))(dQ)/((R^(2)+x^(2)))costheta`
`rArr F_(x)=(1)/(4piepsilon_(0))(Qqx)/((R^(2)+x^(2))^(3//2))`
`a=(1)/(4piepsilon_(0)m)(Qqx)/((R^(2)+x^(2))^(3//2))`
`(d)/(dx)[(x)/((R^(2)+x^(2))^(3//2))]=0`
`rArr (1)/((R^(2)+x^(2))^(3//2))(1)=x(-3)/(2)(R^(2)+x^(2))^(-5//2)(2x)=0`
`rArr (1)/(R^(2)+x^(2))^(3//2)=(3x^(2))/((R^(2)+x^(2))^(5//2)) rArr R^(2)+x^(2)=3x^(2)`
`rArr a_("max")=(1)/(4piepsilon_(0))(QqR//sqrt(2))/((R^(2)+(R^(2))/(2))^(3//2))`
`=(1)/(4piepsilon_(0)mR^(2))(Qq(2sqrt(2)))/(sqrt(2)3sqrt(3))=(Qq)/(6sqrt(3)piepsilon_(0)mR^(2))`
Applying mecharnical energy cosnervation
`(1)/(4piepsilon_(0))(Qq)/(R)=(1)/(4piepsilon_(0))(Qq)/(sqrt(R^(2)+x^(2)))+(1)/(2)mv^(2)`
`KE=(1)/(2)mv^(2)=(1)/(4pepsilon_(0))(Qq)/(R)(1-sqrt((2)/(3)))`
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