2particles having charges q1=0.500 nc and q2=8.00 nc are separated by a distance of 1.20m. at what point along the line connecting the two charges is the total electric field due to the two chares equal to zero?

Refer to the figure shown below.

Charge q₁ = 0.5 nC = 0.5x10⁻⁹ C
Charge q₂ = 8 nC = 8x10⁻⁹ C
d = 1.2 m, the distance between the two charges.

x is the distance between the two charges, measured from the charge q₁.

From Coulomb's Law,
The electric field generated along x by q₁ is
E₁ = k(q₁/x²)
The electric field generated along x by q₂ is
E₂ = -k[q₂/(d-x)²]
where
k = 8.988x10⁹ (N-m²)/C² is the Coulomb constant/

When the electric field along x is zero, then
E₁ + E₂ = 0
k[q₁/x² - q₂/(d-x)²] = 0

That is,
0.5/x² = 8/(1.2 - x)²
8x² = 0.5(1.2 - x)²
16x² = 1.44 - 2.4x + x²
15x² + 2.4x - 1.44 = 0

Solve with the quadratic formula.
x = (1/30)*[-2.4 +/- √(5.76 + 86.4)]
   = 0.24 or -0.4 m

Reject the negative answer to obtain
x = 0.24 m
d-x = 0.96 m

Answer
The electric field is zero between the charges so that
(a) It is at 0.24 m from the 0.5 nC charge, and
(b) It is at 0.96 m from the 8 nC charge.

Letter a, because when there are unlike charges they will always attract to each other.