Two point charges `-q and +q` are located at points `(0,0,-a) and (0,0,a)` respectively. The potential at a point `(0,0,z)`, where `z gt a` is

Options

A. The plot below represents schematically the variation of beat frequency with time
B. `v_(P)+v_(R)=2v_(Q)`
C. The plot below represents schematically the variation of beat frequency with time
D. The rate of change in beat frequency is maximum when the car passes through Q

Answer

A, B, D

Explanation

Sped of car, `V=60 km//h`
`(500)/(3) m//s`
At a point S,between P & Q
`v_m=v_m((c+v costheta)/(c)),v_n=v_n((v+v costheta)/(c))`
`implies trianglev=(v_n-v_n)(1-(v costheta)/(c))`
similarly between Q & R
`implies trianglev=(v_n-v_n)(1-(v costheta)/(c))`
`(d (trianglev))/dt=pm(v_n-v_n)v/csintheta(d theta)/dt`
`theta approx 0^(@)` at P & R as they are large distance apert
`implies` slope of graph is zero
at Q,`theta=90^(@)`
`implies sin theta` is max
also valued of `(d theta)/(dt)` is max
as `(d theta)/(dt)=v/x` where V is its velocity and r is length of kine joining P & S
and r is minimum at Q
`implies` slope is maximum at Q
`implies` (A) & (B) are correct
At p,`triangle v=(v_n-v_n)(1+(v)/(c)) ( theta=0^(@)`
At R, `triangle v=(v_n-v_n)(1-(v)/(c)) ( theta=0^(@)`
At Q, `triangle v=(v_n-v_n) (theta=90^(@))`
`implies trianglev_p+trianglev_R=2(v_n-v_n)=2trianglev_Q` Hence (B) is correct.