The force of repulsion between two like-charged particles will increase if the distance between the charged particles is decreased. According to Coulomb's Law, the force \( F \) between two charges is inversely proportional to the square of the distance \( r \) between them, expressed as:

\[ F = k \frac{{|q_1 \cdot q_2|}}{{r^2}} \]

where \( k \) is Coulomb's constant and \( q_1 \) and \( q_2 \) are the magnitudes of the charges.

Therefore, decreasing the distance \( r \) increases the force of repulsion.

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Question

The force of repulsion between two like-charged particles will increase if the distance between the charged particles is decreased. According to Coulomb's Law, the force \( F \) between two charges is inversely proportional to the square of the distance \( r \) between them, expressed as:

\[ F = k \frac{{|q_1 \cdot q_2|}}{{r^2}} \]

where \( k \) is Coulomb's constant and \( q_1 \) and \( q_2 \) are the magnitudes of the charges.

Therefore, decreasing the distance \( r \) increases the force of repulsion.

GPT-4o mini · Answer

The force of repulsion between two like-charged particles will increase if the distance between the charged particles is decreased. According to Coulomb's Law, the force \( F \) between two charges is inversely proportional to the square of the distance \( r \) between them, expressed as:

\[ F = k \frac{{|q_1 \cdot q_2|}}{{r^2}} \]

where \( k \) is Coulomb's constant and \( q_1 \) and \( q_2 \) are the magnitudes of the charges.

Therefore, decreasing the distance \( r \) increases the force of repulsion.

Answer

witch one is it

GPT-4o mini · Answer

The correct response is: **The distance between the charged particles is decreased.** This is because decreasing the distance between two like-charged particles will increase the force of repulsion between them, according to Coulomb's Law.