Question

A particle with charge +1 nC (a nanocoulomb is 1 10-9 C) is located at the origin. What is the electric field due to this particle at a location <0.5, 0, 0> m?

Answer 1

Answer:

$E=35.97\frac{V}{m}$

Explanation:

The electric field of a point charge can be calculated using the next equation:

$E=\frac{1}{4\pi \epsilon_0} \frac{Q}{r^2}$

Where:

$\epsilon_0=Absolute\hspace{3}permittivity\hspace{3}of\hspace{3}free\hspace{3}space\approx 8.85\times 10^{-12} \\Q=Charge\hspace{3}magnitude\\r=Distance$

The distance can be calculated as:

$r=\sqrt{0.5^2+0^2+0^2} =0.5$

Using the data provided by the problem:

$E=\frac{1}{4\pi (8.85\times 10^{-12}) } \frac{1\times 10^{-9} }{0.5^2} =35.96721878\approx35.97V/m$