Question

Two point charges, 3.0

Answer 1

Consider Point P at x<0 and dist r

EP = E1 + E2

E = kq / r^2

E1 = 9*10^9*(-2)*10^-6 / r^2

E2 = 9*10^9*(3)*10^-6 / (r+6)^2

EP = 0

so , 2/r^2 = 3/(r+6)^2

r = 26.69 , -2.697

Consider Point P at 0<x<6 and dist r from origin

EP = E1 + E2

E = kq / r^2

E1 = 9*10^9*(-2)*10^-6 / r^2

E2 = 9*10^9*(3)*10^-6 / (6-r)^2

EP = 0

so , 2/r^2 = 3/(6-r)^2

r = -26.69 , 2.697

so,

at x < 0 => -2.697

0<x<6 => 0 (no point)

x>6 => 26.69

2)

for Potential also same as E

at x < 0 => -2.697

0<x<6 => 0 (no point)

x>6 => 26.69