Answer :
To determine the electric field intensity at the location of the test charge, we use the relationship defined by Coulomb's Law for electric fields. The electric field intensity \( E \) can be calculated using the formula:
[tex]\[ E = \frac{F}{q} \][/tex]
where:
- \( F \) is the force experienced by the test charge
- \( q \) is the magnitude of the test charge
Given in the problem:
- The force \( F \) on the test charge is \( 0.04 \) Newtons (\( N \)).
- The test charge \( q \) is \( 1.0 \times 10^{-6} \) Coulombs (\( C \)).
Now substituting these values into the formula:
[tex]\[ E = \frac{F}{q} = \frac{0.04 \, \text{N}}{1.0 \times 10^{-6} \, \text{C}} \][/tex]
To simplify this, we perform the division:
[tex]\[ E = \frac{0.04}{1.0 \times 10^{-6}} = \frac{0.04}{0.000001} = 40000 \, \text{N/C} \][/tex]
Thus, the electric field intensity at the location of the test charge is:
[tex]\[ 4.0 \times 10^4 \, \text{N/C} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{4.0 \times 10^4 \, \text{N/C}} \][/tex]
[tex]\[ E = \frac{F}{q} \][/tex]
where:
- \( F \) is the force experienced by the test charge
- \( q \) is the magnitude of the test charge
Given in the problem:
- The force \( F \) on the test charge is \( 0.04 \) Newtons (\( N \)).
- The test charge \( q \) is \( 1.0 \times 10^{-6} \) Coulombs (\( C \)).
Now substituting these values into the formula:
[tex]\[ E = \frac{F}{q} = \frac{0.04 \, \text{N}}{1.0 \times 10^{-6} \, \text{C}} \][/tex]
To simplify this, we perform the division:
[tex]\[ E = \frac{0.04}{1.0 \times 10^{-6}} = \frac{0.04}{0.000001} = 40000 \, \text{N/C} \][/tex]
Thus, the electric field intensity at the location of the test charge is:
[tex]\[ 4.0 \times 10^4 \, \text{N/C} \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{4.0 \times 10^4 \, \text{N/C}} \][/tex]
