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

Direction: Negative x-direction

Step-by-step 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.

Answer 2

The value is
`E =  62 \  N/C`

Step-by-step 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 * (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  = (9*10^9 *  1.60*10^(-19))/( (4.82 *10^(-6)))`

`E =  62 \  N/C`