Question

A rod 8 cm long is uniformly charged and has a total charge of −21.8 μC. The Coulomb constant is 8.98755×10 N·m /C . Determine the magnitude of the electricfield along the axis of the rod at a point 42.279 cm from the center of the rod. Answer in units of N/C.

Answer 1

Answer:

The magnitude of the electric field along the axis of the rod is 1.106 X 10⁶ N/C

Explanation:

The magnitude of the electric field along the axis of a rod is given as;

$E = \frac{\lambda}{4\pi \epsilon_o }[\frac{1}{a} -\frac{1}{L+a}]$

where;

λ is linear charge density, C/m

a is the distance from center of the rod, = (0.42279 - 0.08/2)m = 0.38279 m

L is the length of the rod,  = 0.08 m

ε is the permittivity of free space = 8.85 x 10⁻¹² C²/Nm²

λ  = Q/L = (21.8 μC)/0.08 = (21.8 * 10⁻⁶C)/0.08 = 2.725 x 10⁻⁴ C/m

1/4πε = k = 8.98755 x 10⁹ Nm²/C²

Solving for E

$E = (2.725 X 10^{-4} )(8.98755 X 10^9)[\frac{1}{0.38279} - \frac{1}{0.38279 +0.08}]\\\\ E= 24.491 X10^5[2.6124-2.1608]\\\\E= 24.491 X10^5[0.4516] = 1.106X10^6 N/C$

Therefore, the magnitude of the electric field along the axis of the rod is 1.106 X 10⁶ N/C