Respuesta :
Answer:
The magnitude of the electric field at the given point is 14.57 N/C and the direction of the electric field is negative x- direction
Explanation:
Given that,
Electron located on the x axis is [tex]x=-8.27\times10^{-6}\ m[/tex]
Electric field at a point on the x axis is [tex]x'= 1.67\times10^{-6}\ m[/tex]
Magnitude of charge [tex]q=1.6\times10^{-19}\ C[/tex]
We need to calculate the distance between the electron and the point
[tex]r = x'-x[/tex]
Put the value into the formula
[tex]r =1.67\times10^{-6}-(-8.27\times10^{-6})[/tex]
[tex]r=1.67\times10^{-6}+8.27\times10^{-6}[/tex]
[tex]r=9.94\times10^{-6}\ m[/tex]
We need to calculate the magnitude of the electric field
Using formula of electric field
[tex]E = \dfrac{kq}{r^2}[/tex]
Put the value into the formula
[tex]E=\dfrac{9\times10^{9}\times1.6\times10^{-19}}{(9.94\times10^{-6})^2}[/tex]
[tex]E=14.57\ N/C[/tex]
The direction of the electric field is negative x- direction because the direction of force on the charge at a given point on the x-axis.
Hence, The magnitude of the electric field at the given point is 14.57 N/C and the direction of the electric field is negative x- direction
