Question

An electron is located on the x ‑axis at x0=−2.05×10−6 m . Find the magnitude and direction of the electric field at x=2.77×10−6 m on the x ‑axis due to this electron.

Answer 1

Answer:

The  value is  $E =  62 \  N/C$

Explanation:

From the question we are told that

   The initial position of the electron  $x_o  =  -2.05*10^{-6} \  m$

     The  position that is  considered $x=  2.77*10^{-6} \  m$

Generally the electric field is mathematically represented as

       $E =  k * \frac{q}{r^2 }$

Here k is the coulombs constant  with value  $k  =  9*10^9  \  \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

r is  the distance covered which is mathematically represented as

         $r =  x- x_o$

        $r =  2.77*10^{-6} -(-2.05*10^{-6})$

         $r =  4.82 *10^{-5}$

So

     $E  = \frac{9*10^9 *  1.60*10^{-19}}{ (4.82 *10^{-6})}$

      $E =  62 \  N/C$

     

Answer 2

Answer:

Direction: Negative x-direction

Explanation:

The question tells you that both lie on the   ‑axis, and the data indicate that the electron is in the negative   ‑direction from the field point ( 0< ). Because the field produced by a negative source charge points toward the source, the field in this case points in the negative   ‑direction.