An object carries a +15.5 uC charge. It is 0.525 m from a -7.25 uC charge.
What is the magnitude of the electric
force on the object?

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

dang that balls went far