Electric Potential Question?

Question

A particle with charge +q is at the origin. A particle with -2q us at x=2.00m on the x-axis.
a) What finite values of x is the elextric field zero?
b) What finite values of x is the electric potential zero?

Answer 1

a) For E= 0

E =E1 +(-E2)
0 = E1 +(-E2)

/E1/ = /E2/
kq1/r1 =kq2/r2

where: q2 = 2q; r1 = x, & r2 = 2 + x

kq1/x ² = k2q/ (x +2)²
1/x² = 2/ (x+2)²
(x+2)² = 2x²
x² + 2(x)2 + 4 = 2x²

x² - 4x - 4 = 0 Solving the quadratic equation: we get,

x = + 4.83m
x = - 0.83m


b) For V = 0
Consider any point x between the two point charges.
V = V1 + V2
0 = kq1/x + kq2/(2-x)
where: q1 = q and q2 = (-2q)

0 = kq/x + k(-2q)/(2-x)
0 = kq/x - k(2q)/(2-x)
k(2q)/(2-x) = kq/x
2/(2-x) = 1/x
2x = (2-x)
3x = 2
x = 2/3 m from q1 where V =0

Solution could be shorter if associated with diagrams, and the equation editor can't be used here....
I know you don't need much of the numerical solution...

Answer 2

a) let x = x, keep a test charge at P (x)
Potential at P = PE of system/test charge
V(x) = kq/x + (-2q)k/(2-x)
V(x) = kq/x + 2q k/(x-2) adjusted -ve in fraction ------- (1)
V (x) = 0 given
2-x = 2x
3x = 2
x = 2/3 = 0.667 m
-------------------------
electric field E (P) = - dV (P)/dx differentiate (1)
E = - kq[- 1/x^2 + 2 (- 1)/(x-2)^2]
E = kq[1/x^2 + 2/(x-2)^2]
E =0
-2 x^2 = (x-2)^2
>>> imaginary solution for inside point
====================
if P is on the left of +q
E = 0 = qk/x^2 - 2kq/(x+2)^2
(x+2)^2 = 2 x^2 = x^2+4x+4
x^2 - 4 x - 4 =0
x = [4+-(16 +16)^1/2]/2
x1 = [4+5.66]/2 = 4.83 m
x2 = - 0.33 m
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it has confused me
typical values of (-2q) and 2m are doing mathematical tricks??