Initial velocity and acceleration of a particles are as shown in the figure. Acceleration vector of particle remain constant. Then radius of curvature of path of particle `:`

Options

A. is 9 m initially
B. is `(9)/(sqrt(3))m` initially
C. will have minimum value of `(9)/(8) m`
D. will have minimum value `(3)/(8) m`

Answer

A, C

Explanation

v at origin `ne =`
`E=(r=2 m)=(K(-q)r)/((R_(1)^(2)+r^(2))^(3//2))+(K.Q.r)/((R_(2)^(2)+r^(2))^(3//2))`
`=K.rq[-(1)/(10^(3//2))+(2 sqrt(2))/(2^(3//2)10^(3//2))]=0`
From origin to r=2, field is towards origin.