Question

An electron is located on the x x ‑axis at x 0 = − 3.33 × 10 − 6 m x0=−3.33×10−6 m . Find the magnitude and direction of the electric field at x = 6.25 × 10 − 6 m x=6.25×10−6 m on the x x ‑axis due to this electron.

Answer 1

-15.67287 N/C

Step-by-step explanation:

`x_0` = Location of electron =
`-3.33* 10^(-6)\ m`

`x` = Location of electric field =
`6.25* 10^(-6)\ m`

e = Charge of electron =
`-1.6* 10^(-19)\ C`

k = Coulomb constant =
`8.99* 10^(9)\ Nm^2/C^2`

Distance between the points is

`r=x-x_0\\\Rightarrow r=6.25* 10^(-6)-(-3.33* 10^(-6))\\\Rightarrow r=9.58* 10^(-6)\ m`

Electric field is given by

`E=(ke)/(r^2)\\\Rightarrow E=(8.99* 10^9* -1.6* 10^(-19))/((9.58* 10^(-6))^2)\\\Rightarrow E=-15.67287\ N/C`

The magnitude of the electric field is 15.67287 N/C

The direction is negative as the electron has negative charge.