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 electric field along the axis of the rod at a point 36.0 cm from its center.

Respuesta :

Answer:

[tex]E=1.58*10^6N/C[/tex]

The direction is toward the negative x axis i.e the Rod

Explanation:

From the question we are told that:

Length [tex]L=14.0cm[/tex]

Total charge [tex]q=-22.0\muC[/tex]

Distance [tex]d=36cm=>0.36m[/tex]

Generally the equation for Electric Field is mathematically given by

 [tex]E=\frac{kq}{d^2}[/tex]

Since Magnitude of E is at 36cm away from center

 [tex]E=\frac{kq}{d^2-(L/2)^2}[/tex]

 [tex]E=\frac{9*10^9*(22*10^6)}{0.36^2-(0.14m/2)^2}[/tex]

 [tex]E=1.58*10^6N/C[/tex]

Therefore with Charge Bearing the Negative sign the direction is toward the negative x axis i.e the Rod