Answer:
The electric field strength is equal to 2.50 newtons per coulomb at a distance of 2 meters.
Explanation:
From Classic Electrostatic Theory, we find that electric field ([tex]E[/tex]), measured in newtons per coulomb, is defined by the following equation:
[tex]E = \frac{F}{q_{O}}[/tex] (1)
Where:
[tex]F[/tex] - Electrostatic force, measured in newtons.
[tex]q_{O}[/tex] - Electric charge, measured in coulombs.
In addition and supposing that phenomena occurs between particles, electrostatic force is modelled after the Coulomb's Law:
[tex]F = \frac{k\cdot q\cdot q_{O}}{r^{2}}[/tex] (2)
Where:
[tex]k[/tex] - Electromagnetic constant, measured in newton-square meters per square coulomb.
[tex]q[/tex], [tex]q_{O}[/tex] - Electric charges, measured in coulomb.
[tex]r[/tex] - Distance between particles, measured in meters.
By applying (2) in (1), we get the following definition of electric field:
[tex]E = \frac{k\cdot q}{r^{2}}[/tex] (3)
Then, we observe that electric field is inversely proportional to the square of the distance. The following relationship is therefore constructed:
[tex]\frac{E_{2}}{E_{1}} = \left(\frac{r_{1}}{r_{2}} \right)^{2}[/tex] (4)
If we know that [tex]E_{1} = 10\,\frac{N}{C}[/tex], [tex]E_{2} = 2.50\,\frac{N}{C}[/tex] and [tex]r_{1} = 1\,m[/tex], then the distance is:
[tex]\left(\frac{E_{2}}{E_{1}} \right)\cdot r_{2}^{2}= r_{1}^{2}[/tex]
[tex]r_{2}^{2}=\left(\frac{E_{1}}{E_{2}} \right)\cdot r_{1}^{2}[/tex]
[tex]r_{2} = r_{1}\cdot \sqrt{\frac{E_{1}}{E_{2}} }[/tex]
[tex]r_{2} = (1\,m)\cdot \sqrt{\frac{10\,\frac{N}{C} }{2.50\,\frac{N}{C} } }[/tex]
[tex]r_{2} = 2\,m[/tex]
The electric field strength is equal to 2.50 newtons per coulomb at a distance of 2 meters.
