Given the details of the problem, let's solve for the charge on particle B using Coulomb's law, which states:
[tex]\[ F = k \frac{q_A q_B}{r^2} \][/tex]
Where:
- [tex]\( F = 5.2 \times 10^{-5} \)[/tex] newtons
- [tex]\( r = 2.4 \times 10^{-2} \)[/tex] meters
- [tex]\( q_A = 7.2 \times 10^{-8} \)[/tex] coulombs
- [tex]\( k = 90 \times 10^9 \)[/tex] newton-meters²/coulomb²
We need to find [tex]\( q_B \)[/tex]. We rearrange the formula to solve for [tex]\( q_B \)[/tex]:
[tex]\[ q_B = \frac{F \cdot r^2}{k \cdot q_A} \][/tex]
Now, let's plug in the given values:
[tex]\[ q_B = \frac{(5.2 \times 10^{-5}) \cdot (2.4 \times 10^{-2})^2}{(90 \times 10^9) \cdot (7.2 \times 10^{-8})} \][/tex]
Let's start by calculating the numerator:
[tex]\[ (5.2 \times 10^{-5}) \cdot (2.4 \times 10^{-2})^2 \][/tex]
[tex]\[ (2.4 \times 10^{-2})^2 = 5.76 \times 10^{-4} \][/tex]
[tex]\[ 5.2 \times 10^{-5} \cdot 5.76 \times 10^{-4} = 3.002 \times 10^{-8} \][/tex]
Next, let's calculate the denominator:
[tex]\[ (90 \times 10^9) \cdot (7.2 \times 10^{-8}) = 648 \times 10^1 = 6.48 \times 10^3 \][/tex]
Now, we divide the numerator by the denominator:
[tex]\[ q_B = \frac{3.002 \times 10^{-8}}{6.48 \times 10^3} \][/tex]
[tex]\[ q_B \approx 4.63 \times 10^{-12} \][/tex]
Given the options:
A. [tex]\( 2.4 \times 10^{-2} \)[/tex] coulombs
B. [tex]\( 4.6 \times 10^{-11} \)[/tex] coulombs
C. [tex]\( 5.2 \times 10^{-11} \)[/tex] coulombs
D. [tex]\( 7.2 \times 10^{-8} \)[/tex] coulombs
The closest value to [tex]\( 4.63 \times 10^{-12} \)[/tex] coulombs is option B: [tex]\( 4.6 \times 10^{-11} \)[/tex] coulombs.
Therefore, the correct answer is:
B. [tex]\( 4.6 \times 10^{-11} \)[/tex] coulombs
