Question

A rod 16.0 cm long is uniformly charged and has a total charge of -21.0 µC. Determine the magnitude and direction of the electric field along the axis of the rod at a point 36.0 cm from its center

magnitude	N/C
direction	---Select--- toward the rod away from the rod

Answer 1

The length of the rod is L = 16 cm = 0.16 m

The total charge is, q = -21 uC = -22 * 10^-6 C

The point where the electric field is to be calculated lies on the axis of the rod at a distance 36 cm from its center.

$\lambda = \frac{q}{L}$

   $\lambda =- \frac{-21*10^{-6} C }{0.16 \ m }$

The magnitude and the direction of the elctric field along the axis of the rod at a distance of 36 cm from its center is

$E = K_e \lambda \int_{a + \frac{L}{2}}^{a - \frac{L}{2}} \frac{dx}{x^2}$ ( Here a = 36 cm )

$E = K_e \lambda \left ( \frac{1}{a - \frac{L}{2}} - \frac{1}{a + \frac{L}{2}} \right )$

$= 9 *10^{9} N/C^2 m^2 * \frac{ 21*10^{-6} C }{0.16 \ m } \left ( \frac{1}{0.36-0.8} - \frac{1}{ 0.36+0.8} \right )$

$= - 3.7 * 10^6 \ \mathrm{N/C }$

Here, the direciton is towards the rod.