Question

A rod 14.0 cm long is uniformly charged and has a total charge of -22.0 μC. Determine the magnitude and direction of the net electric field produced by the charged rod at a point 36.0 cm to the right of its center along the axis of the rod.

Answer 1

Explanation:

It is given that,

Length of the rod, l = 14 cm = 0.14 m

Charge on the rod, $q=-22\ \mu C=-22\times 10^{-6}\ C$

We need to find the magnitude and direction of the net electric field produced by the charged rod at a point 36.0 cm to the right of its center along the axis of the rod, z = 36 cm = 0.36 m

Electric field at the axis of the rod is given by :

$E=\dfrac{\lambda}{2\pi \epsilon_o z}$

Where

$\lambda$ is the linear charge density of the rod,

$\lambda=\dfrac{q}{l}=\dfrac{-22\times 10^{-6}\ C}{0.14\ m}=-0.00015\ C/m$

$E=\dfrac{-0.00015}{2\pi\times 8.85\times 10^{-12}\times 0.36}$

E = -7493170.57 N/C

or

$E=-7.49\times 10^6\ N/C$

Negative sign shows that the electric field is acting in inwards direction. Hence, this is the required solution.