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At a distance r1 from a point charge, the magnitude of the electric field created by the charge is 300 N/C. At a distance r2 from the charge, the field has a magnitude of 160 N/C. Find the ratio r2/r1.

Respuesta :

onyebuchinnaji

Answer:

the  ratio of r₂/r₁ is 1.37.

Explanation:

Given;

magnitude of electric field strength at a distance r₁, E₁ = 300 N/C

magnitude of electric field strength at a distance r₂, E₂ = 160 N/C

Electric field strength is given as;

[tex]E = \frac{kQ}{r^2} \\\\E_1r_1^2 = E_2r_2 ^2\\\\\frac{E_1}{E_2} = \frac{r_2^2}{r_1^2} \\\\\frac{E_1}{E_2} =( \frac{r_2}{r_1})^2\\\\\sqrt{\frac{E_1}{E_2} } = \frac{r_2}{r_1}\\\\\frac{r_2}{r_1} = \sqrt{\frac{300}{160} }\\\\\frac{r_2}{r_1} =1.37[/tex]

Therefore, the  ratio of r₂/r₁ is 1.37.

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