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

Answer:

-15.67287 N/C

Explanation:

$x_0$ = Location of electron = $-3.33\times 10^{-6}\ m$

$x$ = Location of electric field = $6.25\times 10^{-6}\ m$

e = Charge of electron = $-1.6\times 10^{-19}\ C$

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Distance between the points is

$r=x-x_0\\\Rightarrow r=6.25\times 10^{-6}-(-3.33\times 10^{-6})\\\Rightarrow r=9.58\times 10^{-6}\ m$

Electric field is given by

$E=\dfrac{ke}{r^2}\\\Rightarrow E=\dfrac{8.99\times 10^9\times -1.6\times 10^{-19}}{(9.58\times 10^{-6})^2}\\\Rightarrow E=-15.67287\ N/C$

The magnitude of the electric field is 15.67287 N/C

The direction is negative as the electron has negative charge.