Question

A charge (uniform linear density = 9.0 nc/m) is distributed along the x axis from x = 0 to x = 3.0 m. determine the magnitude of the electric field at a point on the x axis with x = 4.0 m.

A) 81 N/C
B) 74 N/C
C) 61 N/C
D) 88 N/C
E) 20 N/C

Answer 1

The magnitude of the electric field at a point on the x-axis with x = 4.0 m is 20 N/C.

Step-by-step explanation:

To determine the magnitude of the electric field at a point on the x-axis with x = 4.0 m, we can use the formula for the electric field due to a linear charge distribution:

E = k * λ / r

Where E is the electric field, k is the Coulomb constant (9 * 10^9 Nm^2/C^2), λ is the linear charge density, and r is the distance from the charge distribution. In this case, the linear charge density is given as 9.0 nc/m (or 9.0 * 10^(-9) C/m) and the distance is 4.0 m.

Plugging in the values, we get:

E = (9 * 10^9 Nm^2/C^2) * (9.0 * 10^(-9) C/m) / 4.0 m

E = 20 N/C

Therefore, the magnitude of the electric field at the point on the x-axis with x = 4.0 m is 20 N/C.