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

Answer:

The magnitude of the electric field is 2.4 N/C

Explanation:

Given that,

Charge = 16 nC

We need to calculate the charge density

Using formula of charge density

$\lambda=\dfrac{q}{l}$

Put the value into the formula

$\lambda=\dfrac{16\times10^{-9}}{4}$

$\lambda=4\times10^{-9}\ C$

We need to calculate the magnitude of the electric field

Using formula of electric field

$dE=\dfrac{k dq}{r^2}$

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

On integration

$E=9\times10^{9}\times4\times10^{-9}\times(\dfrac{1}{(10-x)})_{0}^{4}$

$E=36\times(\dfrac{1}{10-4}-\dfrac{1}{10-0})$

$E=36\times(\dfrac{1}{6}-\dfrac{1}{10})$

$E=2.4\ N/C$

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