Answer:
Electric field strength = 17.78 N/C
Explanation:
To find the electric field strength at 0.45 m away from the charge, we can compare it with the electric field strength at 0.15 m.
The formula of electric field strength:
[tex]\boxed{E=\frac{kq}{d^2} }[/tex]
where:
Given:
[tex]\displaystyle E_1:E_2=\frac{kq_1}{d_1^2} :\frac{kq_2}{d_2^2}[/tex]
[tex]\displaystyle 160:E_2=\frac{1}{0.15^2} :\frac{1}{0.45^2}[/tex]
[tex]\displaystyle 160:E_2=0.45^2 :0.15^2[/tex]
[tex]\displaystyle E_2=\frac{0.15^2\times160}{0.45^2}[/tex]
[tex]\displaystyle E_2=\frac{160}{9}[/tex]
[tex]\bf E_2\approx17.78\ N/C[/tex]
