Question

An electron is located on the x axis at x0 = -8.27 × 10-6 m. Find the magnitude and direction of the electric field at x = 1.67 × 10-6 m on the x axis due to this electron. Magnitude:

Answer 1

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

Step-by-step explanation:

Given that,

Electron located on the x axis is
`x=-8.27*10^(-6)\ m`

Electric field at a point on the x axis is
`x'= 1.67*10^(-6)\ m`

Magnitude of charge
`q=1.6*10^(-19)\ C`

We need to calculate the distance between the electron and the point

`r =  x'-x`

Put the value into the formula

`r =1.67*10^(-6)-(-8.27*10^(-6))`

`r=1.67*10^(-6)+8.27*10^(-6)`

`r=9.94*10^(-6)\ m`

We need to calculate the magnitude of the electric field

Using formula of electric field

`E = (kq)/(r^2)`

Put the value into the formula

`E=(9*10^(9)*1.6*10^(-19))/((9.94*10^(-6))^2)`

`E=14.57\ N/C`

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