To solve this problem, we must ensure that the net force on the charge \( q \), when positioned between the charges \( +e \) and \( +2e \), is zero, meaning the system remains in equilibrium. The relevant physics concept here is the equilibrium of electrostatic forces.
The force between two point charges is given by Coulomb's Law:
\(F = \frac{k \times |q_1 \times q_2|}{r^2}\)
where \( F \) is the magnitude of the force, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges.
\(F_1 = \frac{k \times (e \times q)}{x^2}\)
\(F_2 = \frac{k \times (2e \times q)}{(16 - x)^2}\)
\(\frac{k \times (e \times q)}{x^2} = \frac{k \times (2e \times q)}{(16 - x)^2}\)
\(\frac{e}{x^2} = \frac{2e}{(16 - x)^2}\)
\(\frac{1}{x^2} = \frac{2}{(16 - x)^2}\)
\((16 - x)^2 = 2x^2\)
\((16 - x)^2 = 2x^2\)
\(16^2 - 32x + x^2 = 2x^2\)
\(256 - 32x + x^2 = 2x^2\)
\(256 - 32x = x^2\)
\(2x^2 - x^2 + 32x - 256 = 0\)
\(x^2 - 32x + 256 = 0\)
\(x = \frac{32 \pm \sqrt{32^2 - 4 \times 1 \times 256}}{2 \times 1}\)
\(x = \frac{32 \pm \sqrt{1024 - 1024}}{2}\)
\(x = \frac{32 \pm \sqrt{0}}{2}\)
\(x = \frac{32}{2}\)
\(x = 6.63 \text{ cm}\)
Therefore, the correct answer is 6.63 cm from \( +e \).
A 10 $\mu\text{C}$ charge is placed in an electric field of $ 5 \times 10^3 \text{N/C} $. What is the force experienced by the charge?