Answer:
67.5 N
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = 3.0×10¯⁶ C
Charge 2 (q₂) = 2.5×10¯⁷ C
Distance apart (r) = 1 cm
Force (F) =?
Next, we shall convert 1 cm to m. This can be obtained as follow:
100 cm = 1 m
Therefore,
1 cm = 1 cm × 1 m / 100 cm
1 cm = 0.01 m
Thus, 1 cm is equivalent 0.01 m.
Finally, we shall determine the electrical force between the two points charges as follow:
Charge 1 (q₁) = 3.0×10¯⁶ C
Charge 2 (q₂) = 2.5×10¯⁷ C
Distance apart (r) = 0.01 m
Electrical constant (K) = 9×10⁹ Nm²C¯²
Force (F) =?
F = Kq₁q₂ / r²
F = 9×10⁹ × 3.0×10¯⁶ × 2.5×10¯⁷ / (0.01)²
F = 0.00675 / 0.0001
F = 67.5 N
Therefore, the electrical force between the two points charges is 67.5 N
