Answer:
3.67 N
Explanation:
From the question given above, the following data were obtained:
Charge of 1st object (q₁) = +15.5 μC
Charge of 2nd object (q₂) = –7.25 μC
Distance apart (r) = 0.525 m
Force (F) =?
Next, we shall convert micro coulomb (μC) to coulomb (C). This can be obtained as follow:
For the 1st object
1 μC = 1×10¯⁶ C
Therefore,
15.5 μC = 15.5 × 1×10¯⁶
15.5 μC = 15.5×10¯⁶ C
For the 2nd object:
1 μC = 1×10¯⁶ C
Therefore,
–7.25 μC = –7.25 × 1×10¯⁶
–7.25 μC = –7.25×10¯⁶ C
Finally, we shall determine the force. This can be obtained as follow:
Charge of 1st object (q₁) = +15.5×10¯⁶ C
Charge of 2nd object (q₂) = –7.25×10¯⁶ C
Distance apart (r) = 0.525 m
Electrical constant (K) = 9×10⁹ Nm²/C²
Force (F) =?
F = Kq₁q₂ / r²
F = 9×10⁹ × 15.5×10¯⁶ × 7.25×10¯⁶ / 0.525²
F = 3.67 N
Therefore, the force on the object is 3.67 N
