To find the electric field intensity at the location of the test charge, we can use the relationship between the electric field (\(E\)), the force (\(F\)), and the charge (\(q\)). The formula for electric field intensity is given by:
[tex]\[ E = \frac{F}{q} \][/tex]
where:
- \(E\) is the electric field intensity,
- \(F\) is the force on the test charge,
- \(q\) is the magnitude of the test charge.
Given values:
- The force (\(F\)) on the test charge is \(0.06 \, \text{N}\).
- The test charge (\(q\)) is \(2.0 \times 10^{-4} \, \text{C}\).
Plugging these values into the formula:
[tex]\[ E = \frac{0.06 \, \text{N}}{2.0 \times 10^{-4} \, \text{C}} \][/tex]
Now, let's perform the division:
[tex]\[ E = \frac{0.06}{2.0 \times 10^{-4}} \][/tex]
[tex]\[ E = \frac{0.06}{0.0002} \][/tex]
To simplify the division, we can convert the denominator into a common fraction:
[tex]\[ E = \frac{0.06}{0.0002} = \frac{0.06}{2 \times 10^{-4}} = \frac{0.06}{2} \times 10^4 \][/tex]
Simplifying the expression further:
[tex]\[ \frac{0.06}{2} = 0.03 \][/tex]
[tex]\[ 0.03 \times 10^4 = 300 \][/tex]
Thus, the electric field intensity is:
[tex]\[ E = 300 \, \text{N} / \text{C} \][/tex]
Hence, the correct answer is:
[tex]\[ \boxed{3.0 \times 10^2 \, \text{N} / \text{C}} \][/tex]
