Question

charged objects a and b exert electric forces on each other.
complete the statements.

• the electric forces are stronger when a and b have

and are

.

Explanation:

To solve this, we recall Coulomb's Law for electric forces, \( F = k\frac{|q_1q_2|}{r^2} \), where \( F \) is the electric force, \( k \) is a constant, \( q_1, q_2 \) are the charges, and \( r \) is the distance between them.

Step 1: Analyze the effect of charge magnitude

From the formula, the force \( F \) is directly proportional to the product of the magnitudes of the charges (\( |q_1q_2| \)). So, larger magnitudes of charge (more charge) will lead to a stronger electric force.

Step 2: Analyze the effect of distance

The force \( F \) is inversely proportional to the square of the distance (\( r^2 \)) between the charges. Thus, a smaller distance (closer together) will lead to a stronger electric force.

So, the electric forces are stronger when \( \mathbf{A} \) and \( \mathbf{B} \) have more charge (larger magnitude of charge) and are closer together (smaller distance apart).

Answer:

To solve this, we recall Coulomb's Law for electric forces, \( F = k\frac{|q_1q_2|}{r^2} \), where \( F \) is the electric force, \( k \) is a constant, \( q_1, q_2 \) are the charges, and \( r \) is the distance between them.

Step 1: Analyze the effect of charge magnitude

From the formula, the force \( F \) is directly proportional to the product of the magnitudes of the charges (\( |q_1q_2| \)). So, larger magnitudes of charge (more charge) will lead to a stronger electric force.

Step 2: Analyze the effect of distance

The force \( F \) is inversely proportional to the square of the distance (\( r^2 \)) between the charges. Thus, a smaller distance (closer together) will lead to a stronger electric force.

So, the electric forces are stronger when \( \mathbf{A} \) and \( \mathbf{B} \) have more charge (larger magnitude of charge) and are closer together (smaller distance apart).