Answer :
To determine how the electric force between two charged particles changes when the distance between them is increased by a factor of 3, we can use Coulomb's law.
Coulomb’s Law states that the electric force (F) between two charged particles is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them. Mathematically, it can be expressed as:
[tex]\[ F \propto \frac{1}{r^2} \][/tex]
This implies that if the initial distance is [tex]\( r \)[/tex], and the distance is increased by a factor of 3, the new distance becomes [tex]\( 3r \)[/tex].
We can now find the new force (F') using the new distance. The new force is inversely proportional to the square of the new distance:
[tex]\[ F' \propto \frac{1}{(3r)^2} \][/tex]
Simplifying the expression inside the parentheses:
[tex]\[ F' \propto \frac{1}{9r^2} \][/tex]
Now, to find out how much the new force has changed in comparison to the old force, we take the ratio of the new force (F') to the old force (F):
[tex]\[ \frac{F'}{F} = \frac{\frac{1}{9r^2}}{\frac{1}{r^2}} \][/tex]
Dividing the fractions:
[tex]\[ \frac{F'}{F} = \frac{1}{9r^2} \times \frac{r^2}{1} = \frac{1}{9} \][/tex]
Therefore, the new force is [tex]\( \frac{1}{9} \)[/tex] of the old force, meaning the force is reduced by a factor of 9.
So the correct answer is:
OC. It is reduced by a factor of 9.
Coulomb’s Law states that the electric force (F) between two charged particles is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance (r) between them. Mathematically, it can be expressed as:
[tex]\[ F \propto \frac{1}{r^2} \][/tex]
This implies that if the initial distance is [tex]\( r \)[/tex], and the distance is increased by a factor of 3, the new distance becomes [tex]\( 3r \)[/tex].
We can now find the new force (F') using the new distance. The new force is inversely proportional to the square of the new distance:
[tex]\[ F' \propto \frac{1}{(3r)^2} \][/tex]
Simplifying the expression inside the parentheses:
[tex]\[ F' \propto \frac{1}{9r^2} \][/tex]
Now, to find out how much the new force has changed in comparison to the old force, we take the ratio of the new force (F') to the old force (F):
[tex]\[ \frac{F'}{F} = \frac{\frac{1}{9r^2}}{\frac{1}{r^2}} \][/tex]
Dividing the fractions:
[tex]\[ \frac{F'}{F} = \frac{1}{9r^2} \times \frac{r^2}{1} = \frac{1}{9} \][/tex]
Therefore, the new force is [tex]\( \frac{1}{9} \)[/tex] of the old force, meaning the force is reduced by a factor of 9.
So the correct answer is:
OC. It is reduced by a factor of 9.
