Answer
a) negative
b) $x=0.25m$
c) $x=-0.5m$
Work Step by Step
(a) We know that
$V_A=Kq(\frac{1}{r}-\frac{3}{r^{\prime}})$
We plug in the known values to obtain:
$V_A=(9\times 10^9Nm^2/C^2)(4.11\times 10^{-9}C)(\frac{1}{0.333m}-\frac{3}{0.667m})=-55.3V$
Thus, the electric potential is negative at $x=0.333m$
(b) We know that
$V_B=K(\frac{q}{x}-\frac{3q}{1-x})$
$0=K(\frac{q}{x}-\frac{3q}{1-x})$
This simplifies to:
$x=0.25m$
The required point is $x=0.25m$
(c) We know that
$v_C=K(\frac{q}{x}-\frac{3q}{1+x})$
$\implies 0=K(\frac{q}{x}-\frac{3q}{1+x})$
This simplifies to:
$x=0.5m$
As the point is on the left of the origin, therefore $x=-0.5m$
