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 `+0.1muC` moves along the x-direction. Its speed at `x=+ infty` is `v_(0)`. find the least value of `v_(0)` for which the particle will cross the origin. find also the kinetic energy of the particle at the origin in tyhis case. Assume that there is no force part from electrostatic force.

Answer

A, C, D

Explanation

In the figure
`q=muC=10^-6C`
`q_0=+0.1muC=10^-7C`
`m=6xx10^-4kg`
and `Q=8muC=8xx10^-6C`
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/(4piepsilon_0)=9xx10^-9Nm^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)(-1/2)(3/2+x^2)^(-3//2)](2x)`
`E=0` on x-axis where
`8/((27/2+x^2)^(3//2))=1/((3/+x^2)^/(3//2)`
`(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 the particle upto `x=+sqrt(5/2)m` because
at `x=+sqrt(5/2)m,E=0implies` Electrosatic force on charge `q_0` is zero or `F_e=0`
for `xltsqrt(5/2)`m, E is is attractive (towards negative x-axis)
Now, from eqn i potential at `x=sqrt(5/2)m`
`V=1.8xx10^4[8/sqrt(27/2+5/2)-1/sqrt(3/2+5/2)`
`V=2.7xx10^4V`
Applying energy conservation at `x=oo` and `x=sqrt(5/2)m`
`1/2mv_0^2=q_0V`
`:. v_0=sqrt((2q_0V)/m)`
Substituting the values `v_0=sqrt((2xx10^-7xx2.7xx10^4)/(6xx10^-4))`
`v_0=3m/s`
`:.` Minimum value of `v_0` is `3/m/s`.
From eqn i potential at origin `(x=0)` is
`V_0=1.8xx10^4[8/(sqrt27/2)-1/sqrt(3/2)]=2.4xx10^4V`
Let T be the kinetic energy of the particle at origin.
Applying energy conservation at `x=0` and at `x=oo`
T+q_0V_0=1/2mv_0^2`
But `1/2mv_0^2=v_0V` [from eqn ii]
`:. T=q_0(V-V_0)`
`T+(10^-7)(2.7xx10^4-2.4xx10^4)`
`T=3xx10^-4J`