Two charges have a force of attraction between them. If you double the distance between the charges what happens to the force? The force is reduced to 1/2 of its value. The force is reduced to 1/4 of its value. The force is doubled. The force is four times what it was.
Two charges have a force of attraction between them. If you double the distance between the charges what happens to the force? The force is reduced to 1/2 of its value. The force is reduced to 1/4 of its value. The force is doubled. The force is four times what it was.
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Question:**
Two charges have a force of attraction between them. If you double the distance between the charges, what happens to the force?
- The force is reduced to 1/2 of its value.
- The force is reduced to 1/4 of its value.
- The force is doubled.
- The force is four times what it was. (Highlighted)
**Explanation:**
This question revolves around Coulomb's Law, which describes the force of attraction or repulsion between two charged objects. According to this law, the force \( F \) between two charges is inversely proportional to the square of the distance \( r \) between them, expressed as:
\[ F = k \frac{{|q_1 q_2|}}{{r^2}} \]
where \( k \) is Coulomb's constant, and \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
If the distance \( r \) is doubled, the new force \( F' \) is:
\[ F' = k \frac{{|q_1 q_2|}}{{(2r)^2}} = k \frac{{|q_1 q_2|}}{{4r^2}} = \frac{F}{4} \]
Thus, the correct answer is that the force is reduced to 1/4 of its original value. The option "The force is reduced to 1/4 of its value" is correct, not "The force is four times what it was."](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F58e34b7a-1a63-4f47-9def-f0666a842414%2Fe54a2d6a-1f58-4db3-bbdc-375452880b75%2Fj1dp8j8_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
Two charges have a force of attraction between them. If you double the distance between the charges, what happens to the force?
- The force is reduced to 1/2 of its value.
- The force is reduced to 1/4 of its value.
- The force is doubled.
- The force is four times what it was. (Highlighted)
**Explanation:**
This question revolves around Coulomb's Law, which describes the force of attraction or repulsion between two charged objects. According to this law, the force \( F \) between two charges is inversely proportional to the square of the distance \( r \) between them, expressed as:
\[ F = k \frac{{|q_1 q_2|}}{{r^2}} \]
where \( k \) is Coulomb's constant, and \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
If the distance \( r \) is doubled, the new force \( F' \) is:
\[ F' = k \frac{{|q_1 q_2|}}{{(2r)^2}} = k \frac{{|q_1 q_2|}}{{4r^2}} = \frac{F}{4} \]
Thus, the correct answer is that the force is reduced to 1/4 of its original value. The option "The force is reduced to 1/4 of its value" is correct, not "The force is four times what it was."
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