Answer:
[tex]E=35.97\frac{V}{m}[/tex]
Explanation:
The electric field of a point charge can be calculated using the next equation:
[tex]E=\frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}[/tex]
Where:
[tex]\epsilon_0=Absolute\hspace{3}permittivity\hspace{3}of\hspace{3}free\hspace{3}space\approx 8.85\times 10^{-12} \\Q=Charge\hspace{3}magnitude\\r=Distance[/tex]
The distance can be calculated as:
[tex]r=\sqrt{0.5^2+0^2+0^2} =0.5[/tex]
Using the data provided by the problem:
[tex]E=\frac{1}{4\pi (8.85\times 10^{-12}) } \frac{1\times 10^{-9} }{0.5^2} =35.96721878\approx35.97V/m[/tex]
