Saved Bookmarks
| 1. |
Four point charges +8mC, -mC, and +8mC 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^-4kg and charge +0.1muC moves along the -x direction. Its speed at x=+oo 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. Assume that space is gravity free. Given (1)/(4piepsilon_0)=9xx10^9Nm^2//C^2. |
|
Answer» <P> Similarly there are two forces of attraction due to two charges of `-1 mC`. The net force due to these force is `2F cos beta` towards left.  The net force on charge `0.1 muC` is zero when `2F cos alpha=2F^' cos beta` `(Kxx8xx10^-6xx0.1xx10^-6)/(sqrt(x^2+27/2))^2xx(x)/(sqrt(x^2+27/2))` `=(Kxx1xx10^-6xx0.1xx10^-6)/(sqrt(x^2+3/2))^2xx(x)/(sqrt(x^2+3/2))` `implies x=+-sqrt(5/2)` This means that we need to move the charge from `-oo` to `sqrt(5/2)` . Thereafter the attractive forces will make the charge move to origin. The electric potential of the four charges at `x=sqrt(5/2)` is `V=(2xx9xx10^9xx8xx10^-6)/(sqrt(5/2)+27/2)`-(2xx9xx10^9xx10^-6)/(sqrt(5/2)+3/2)` `=2xx9xx10^9xx10^-6[8/4-1/2]=2.7xx10^4V` Kinetic energy is required to overcome the force of repulsion from `prop` to `x=sqrt(5/2)`. The work DONE in this process is `W=q(V)` where V=p.d between `oo` and `x=sqrt(5/2)`. `:.` `W=0.1xx10^-6xx2.7xx10^4=2.7xx10^-3J` By energy conservation `1/2mV_0^2=2.7xx10^-3` `implies 1/2xx6xx10^-4V_0^2=2.7xx10^-3` `implies V_0=3m//s` K.E. at the origin Potential at origin `V_(x=0)=(2xx9xx10^9xx8xx10^-6)/(sqrt(27/2))-(2xx9xx10^9xx10^-6)/(sqrt(3/2))` `=2.4xx10^4` Again by energy conservation `K.E. =q[V_(x=(sqrt5)/(2))-V_(x=0)]` `:.` K.E. =0.1xx10^-6[2.7xx10^4-2.4xx10^4]` `=0.1xx10^-6xx0.3xx10^4` `=3xx10^-4J` |
|