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The force of attraction between two like-charged table tennis balls is [tex]\(2.4 \times 10^{-5}\)[/tex] newtons. If the charge on one ball is [tex]\(3.8 \times 10^{-8}\)[/tex] coulombs and on the other is [tex]\(3.0 \times 10^{-8}\)[/tex] coulombs, what is the distance between the two charges? [tex]\((k = 9.0 \times 10^9 \, \text{N m}^2/\text{C}^2)\)[/tex]

A. 0.11 meters
B. 0.24 meters
C. 0.45 meters
D. 0.65 meters


Sagot :

To solve for the distance between the two like-charged table tennis balls, we can use Coulomb's law. Coulomb's law states that the magnitude of the force [tex]\( F \)[/tex] between two point charges [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] separated by a distance [tex]\( r \)[/tex] is given by:

[tex]\[ F = k \frac{q_1 q_2}{r^2} \][/tex]

where:
- [tex]\( F \)[/tex] is the magnitude of the force between the charges,
- [tex]\( k \)[/tex] is Coulomb's constant ( [tex]\( 9.0 \times 10^9 \)[/tex] N m²/C²),
- [tex]\( q_1 \)[/tex] and [tex]\( q_2 \)[/tex] are the magnitudes of the charges,
- [tex]\( r \)[/tex] is the distance between the charges.

Given:
- [tex]\( F = 2.4 \times 10^{-5} \)[/tex] N
- [tex]\( q_1 = 3.8 \times 10^{-8} \)[/tex] C
- [tex]\( q_2 = 3.0 \times 10^{-8} \)[/tex] C
- [tex]\( k = 9.0 \times 10^9 \)[/tex] N m²/C²

We need to solve the formula for [tex]\( r \)[/tex]. Rearrange Coulomb's law to solve for [tex]\( r \)[/tex]:

[tex]\[ r^2 = k \frac{q_1 q_2}{F} \][/tex]

Taking the square root of both sides:

[tex]\[ r = \sqrt{\frac{k q_1 q_2}{F}} \][/tex]

Now substitute the given values into the equation:

[tex]\[ r = \sqrt{\frac{(9.0 \times 10^9) (3.8 \times 10^{-8}) (3.0 \times 10^{-8})}{2.4 \times 10^{-5}}} \][/tex]

Solving this gives us:

[tex]\[ r \approx 0.653834841531101 \][/tex]

Therefore, the distance [tex]\( r \)[/tex] is approximately 0.65 meters. Looking at the answer choices:
A. 0.11 meters
B. 0.24 meters
C. 0.45 meters
D. 0.65 meters

The correct answer is:
D. 0.65 meters