Question

A negative charge of -2.0 C and a positive charge of 3.0 C are separated by 80 m. What is the electrostatic force between the two charges? Is the force attractive or repulsive? How do you know?

Answer 1

Answer:

1. 8437500 N

2. The force between the two charges is attractive.

Explanation:

1. Determination of the force between the two charges.

Charge 1 (q₁) = –2.0 C

Charge 2 (q₂) = 3.0 C

Distance apart (r) = 80 m

Electrical constant (K) = 9×10⁹ Nm²/C²

Force (F) =?

F = Kq₁q₂ / r²

F = 9×10⁹ × 2 × 3 / 80²

F = 5.4×10¹⁰ / 6400

F = 8437500 N

Thus, the force of attraction between the two charges is 8437500 N

2. From the question given, the charges are:

Charge 1 (q₁) = –2.0 C

Charge 2 (q₂) = 3.0 C

We understood that like charges repels while unlike charges attract. Since the two charges (i.e –2 C and 3 C) has opposite signs, it means they will attract each other.

Thus the force between them is attractive.