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Four point charges +8muC-1muC,-1muC and ...

Four point charges `+8muC-1muC,-1muC and +8muC` are fixed at the points `-sqrt(27//2)m, -sqrt(3//2)m+sqrt(3//2)m and + sqrt(27//2)m` respectively on the Y-axis. A particle of mass `6xx10^-4` kg and charge `+0.1muC` moves along the -X direction. It speed at `x=+oo` is `v_0`. Find the least value of `v_0` for which the paticle will cross the origin. Find also the kinetic energy of the particle at the origin. Assume that space is gravity free.

Text Solution

Verified by Experts

The correct Answer is:
`(v_(0))_("min")=3 m//s, K=3xx10^(-4) J`
In the figure

`q=1 muC=10^(-6)C, q_(0)=+0.1 muC=10^(-7) C`
and
`m=6xx10^(-4) kg` and `Q=8 muC=8cc10^(-6) C`
Let P be any point at a distance x from origin O. Then
`AP=CP=sqrt(3/2+x^(2))`
`BP=DP=sqrt(27/2+x^(2))`
Electric potential at point P will be
`V=(2KQ)/(BP)-(2Kq)/(AP)`
where `K=1/(4pi epsilon_(0))=9xx10^(9) N//m^(2)//C^(2)`
`:. V=2xx9xx10^(9)[(8xx10^(-6))/sqrt(27/2+x^(2))-10^(-6)/sqrt(3/2+x^(2))]`
`V=1.8xx10^(4)[8/sqrt(27/2+x^(2))-1/sqrt(3/2+x^(2))]` ...(i)
`:.` Electric field at P is
`E=-(dV)/(dx)=1.8xx10^(4)`
`[(8)((-1)/2)(27/2+x^(2))^(-3/2)-(-1/2)(3/2+x^(2))^(-3/2)](2x)`
`E=0` on x-axis where `-x=0` or
`8/((27/2+x^(2))^(3//2))=(1)/((3/2+x^(2))^(3//2))`
`rArr ((4)^(3//2))/((27/2+x^(2))^(3//2))=1/((3/2+x^(2))^(3//2))`
`rArr (27/2+x^(2))=4(3/2+x^(2))`
This equation gives `x=+-sqrt(5/2)m`
The least value of kinetic energy of the particle at infinity should be enough to take particle upto
`x=+ sqrt(5/2)m` because at `x=+sqrt(5/2)m, E=0`
`rArr` Electrotatic force on charge q is zero or `F_(e)=0`. For at `x gt sqrt(5/2)m`, E is repulsive (towards x-axis) and for `x lt sqrt(5/2) m`, E is attractive (towards negative x-axis).
Now, from Eq. (i) potential at `x=sqrt(5/2) m`
`v=1.8xx10^(4)[8/sqrt(27/2+5/2)-1/sqrt(3/2+5/2)]`
`rArr V=2.7xx10^(4)` volt
Applying energy conservation at `x=oo` and `x=sqrt(5/2)m, 1/2 mv_(0)^(2)=q_(0)V` ...(ii)
`:. v_(0)=sqrt((2q_(0)V)/m)`
Substituting the values
`v_(0)=sqrt((2xx10^(-7)xx2.7xx10^(4))/(6xx10^(-4)))rArr v_(0)=3 m//s`
`:.` Minimum value of `v_(0)` is `3 m//s`
From eq. (i) potential at origin `(x=0)` is
`V_(0)=1.8xx10^(4)[8/sqrt(27/2)-1/sqrt(3/2)]~~2.4xx10^(4) V`
Let K be the kinetic energy of the particle at origin. Applying energy conservation at x=0 and `x=oo`
`K+q_(0)V_(0)=1/2 mv_(0)^(2)` But `1/2 mv_(0)^(2)=q_(0)V` [from Eq. (ii)]
`K=q_(0)(V-V_(0))=(10^(-7))(2.7xx10^(4)-2.4xx10^(4))`
`rArr K=3xx10^(-4) J`
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