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charge of uniform density 4.0 nc/m is distributed along the x axis from x = -2.0 m to x = 3.0 m. what is the magnitude of the electric field at the point x = 5.0 m on the x axis?

User ZiviMagic
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1 Answer

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Final answer:

To find the magnitude of the electric field at x = 5.0 m on the x-axis, we can use the formula for the electric field. Using the given information, we calculate the electric field as 36.0 N/C.

Step-by-step explanation:

To find the magnitude of the electric field at the point x = 5.0 m on the x-axis, we can use the formula for the electric field created by a charged rod:

E = k * (total charge)/r^2

In this case, the total charge is the charge per meter multiplied by the length of the rod, and r is the distance from the charge to the point where we are calculating the electric field. Using the given information, we can calculate the electric field at x = 5.0 m.

First, we need to find the length of the rod, which is 3.0 m - (-2.0 m) = 5.0 m. Next, we calculate the total charge: charge per meter * length = 4.0 nC/m * 5.0 m = 20.0 nC.

Finally, we can plug these values into the formula:

E = (9.0 * 10^9 N*m^2/C^2) * (20.0 * 10^-9 C) / (5.0 m)^2

E = 36.0 N/C

Therefore, the magnitude of the electric field at the point x = 5.0 m on the x-axis is 36.0 N/C.

User Jimmy Hannon
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