Question

2. 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 rod, l = 14 cm = 0.14 m

Total charge, $Q=-22\ \mu C=-22\times 10^{-6}\ C$

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 rod is given by :

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

Where

$\lambda$ is the linear charge density, $\lambda=\dfrac{Q}{l}$

So, $E=\dfrac{Q}{2\pi \epsilon_o zl}$

$E=\dfrac{-22\times 10^{-6}}{2\pi \times 8.85\times 10^{-12}\times 0.36\times 0.14}$

E = −7849988.22 N/C

or

$E=-7.84\times 10^6\ N/C$

Negative sign shows the direction of electric field is inward in all direction. Hence, this is the required solution.