Question

A 16-nC charge is distributed uniformly along the x axis from x = 0 to x = 4 m. Write down the integral to calculate the magnitude (in N/C) of the electric field at x = +10 m on the x axis?

Answer 1

The magnitude of the electric field is 2.4 N/C

Step-by-step explanation:

Given that,

Charge = 16 nC

We need to calculate the charge density

Using formula of charge density

`\lambda=(q)/(l)`

Put the value into the formula

`\lambda=(16*10^(-9))/(4)`

`\lambda=4*10^(-9)\ C`

We need to calculate the magnitude of the electric field

Using formula of electric field

`dE=(k dq)/(r^2)`

`dE=\int_(0)^(4){(kdq)/((10-x)^2)}`

On integration

`E=9*10^(9)*4*10^(-9)*((1)/((10-x)))_(0)^(4)`

`E=36*((1)/(10-4)-(1)/(10-0))`

`E=36*((1)/(6)-(1)/(10))`

`E=2.4\ N/C`

Hence, The magnitude of the electric field is 2.4 N/C