C

Step 1: Identify the charges.
The charges are arranged at positions 1 m, 2 m, ..., 20 m on the X-axis. We can observe that the charges increase following a specific pattern.
Charge at position x = n m is given by:
$Q_n = n^3 \times 10^{-6} C$ for n = 1, 2, 3,...,20.
Therefore:
Q_1 = 1.0 × 10^(-6) C,
Q_2 = 8.0 × 10^(-6) C,
Q_3 = 27.0 × 10^(-6) C,
Q_4 = 64.0 × 10^(-6) C,...
Q_{20} = 8000 × 10^(-6) C.

Step 2: Calculate the electric force on the 1 C charge.
The electric force due to a charge Q at distance r is given by Coulomb's Law:
$F = k \frac{|Q| \cdot |q|}{r^2}$,
where k = $8.99 × 10^9 \frac{N \cdot m^2}{C^2}$, and q = 1 C is the charge at the origin.
Each of the 20 charges generates a force on the 1 C charge placed at the origin. Since all charges are positive, all forces will be repulsive.
The total force will be the vector sum of the forces from each charge. However, since the origin lies at 0, we can compute the individual forces as they all act in the positive X-direction.

• For $n = 1$: $F_1 = k \frac{1 \times 10^{-6} \cdot 1}{1^2} = 8.99 × 10^3 N$
• For $n = 2$: $F_2 = k \frac{8 \times 10^{-6} \cdot 1}{2^2} = 8.99 × 10^3 N$
• For $n = 3$: $F_3 = k \frac{27 \times 10^{-6} \cdot 1}{3^2} = 8.99 × 10^3 N$
• 7 and so on until n=20.

Step 3: Sum up all the forces.
Total force $F_{total} = F_1 + F_2 + F_3 + ... + F_{20}$.
Calculate the total force.

Therefore, we find the total force acting on the 1C charge placed at the origin is approximately 439,500 N.