To determine the distance from the ball's center where the electric field strength is [tex]\( 5.8 \times 10^5 \)[/tex] newtons/coulomb given a charge [tex]\( q = 1.5 \times 10^{-9} \)[/tex] coulombs and Coulomb's constant [tex]\( k = 9.0 \times 10^9 \)[/tex] newton [tex]\(\cdot\)[/tex] meter[tex]\(^2\)[/tex] / coulomb[tex]\(^2\)[/tex], we can use the formula for the electric field due to a point charge, which is:
[tex]\[ E = \frac{k \cdot q}{r^2} \][/tex]
We need to find the distance [tex]\( r \)[/tex]. Rearranging the formula to solve for [tex]\( r \)[/tex], we get:
[tex]\[ r^2 = \frac{k \cdot q}{E} \][/tex]
Plugging in the values [tex]\( k = 9.0 \times 10^9 \)[/tex] N[tex]\(\cdot\)[/tex]m[tex]\(^2\)[/tex]/C[tex]\(^2\)[/tex], [tex]\( q = 1.5 \times 10^{-9} \)[/tex] C, and [tex]\( E = 5.8 \times 10^5 \)[/tex] N/C, we have:
[tex]\[ r^2 = \frac{(9.0 \times 10^9) \cdot (1.5 \times 10^{-9})}{5.8 \times 10^5} \][/tex]
Calculating the numerator first:
[tex]\[ 9.0 \times 10^9 \times 1.5 \times 10^{-9} = 13.5 \][/tex]
So we get:
[tex]\[ r^2 = \frac{13.5}{5.8 \times 10^5} \][/tex]
Dividing 13.5 by [tex]\( 5.8 \times 10^5 \)[/tex]:
[tex]\[ r^2 = \frac{13.5}{580000} = 2.327586206896552 \times 10^{-5} \][/tex]
To find [tex]\( r \)[/tex], take the square root of both sides:
[tex]\[ r = \sqrt{2.327586206896552 \times 10^{-5}} \][/tex]
Taking the square root:
[tex]\[ r \approx 0.004824506406770077 \][/tex]
Thus, the distance [tex]\( r \)[/tex] is approximately [tex]\( 0.0048 \)[/tex] meters, or [tex]\( 4.8 \times 10^{-3} \)[/tex] meters.
So, the correct answer is:
B. [tex]\( 4.8 \times 10^{-3} \)[/tex] meters
