Answer:
Given the conditions:
The magnitude of the electric field is 220 N/C
The distance moved by the charge in the direction of the electric field is 2.0 cm
The change in the electrical potential energy is 28.82 × 10^-19 J
We can use the equation for the change in electrical potential energy to find the magnitude of the charge on the moving particle:
ΔV = -qEΔx
28.82 × 10^-19 J = -q(220 N/C)(2.0 cm)
To solve for q, we can rearrange the equation and divide both sides by the magnitude of the electric field:
q = (28.82 × 10^-19 J) / (220 N/C * 2.0 cm)
q = 1.301 × 10^-19 C
Therefore, the magnitude of the charge on the moving particle is 1.301 × 10^-19 C
