To solve this problem we will apply the concepts related to the electric field. This definition is based on the proposal by Coulomb and will be described in its two states. Under this relationship we will obtain the radius of the distances depending on the elective field.
The electric field strength of the charge from a distance [tex]r_1[/tex] is,
[tex]E_1 = k \frac{Q}{r_1^2}[/tex]
The electric field strength of the charge from a distance [tex]r_2[/tex] is,
[tex]E_2 = k \frac{Q}{r_2^2}[/tex]
The relation between the electric field strengths [tex]E_1[/tex] and [tex]E_2[/tex] is
[tex]\frac{E_1}{E_2}[/tex] and can be written as
[tex]\frac{E_1}{E_2} = \frac{k \frac{Q}{r_1^2}}{k \frac{Q}{r_2^2}}[/tex]
[tex]\frac{E_1}{E_2} = \frac{r_2^2}{r_1^2}[/tex]
[tex]\frac{r_2}{r_1} = \sqrt{\frac{E_1}{E_2}}[/tex]
Replacing we have,
[tex]\frac{r_2}{r_1} = \sqrt{\frac{344N/C}{177N/C}}[/tex]
[tex]\frac{r_2}{r_1} = 1.39[/tex]
The ratio of the distances [tex]r_2[/tex] and [tex]r_1[/tex] is 1.39
The ratio of [tex]\frac{r_{2}}{r_{1}}[/tex] is equal to 2.01
The Electric field is inversely proportional to distance from charge.
[tex]\frac{E_{1}}{E_{2}} =\frac{r_{2}}{r_{1}}[/tex]
Given that, [tex]E_{1}=344N/C,E_{2}=171N/C[/tex]
Substitute values in above relationship.
[tex]\frac{r_{2}}{r_{1}}=\frac{344}{171} =2.01[/tex]
Thus, The ratio of [tex]\frac{r_{2}}{r_{1}}[/tex] is equal to 2.01
Learn more:
https://brainly.com/question/11509296
