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The magnitude of the electric field due to a small charged object is 12.0 n/c at a distance of 301 cm from the charge. the field 6.00 m away from the charge is. a. 4.0N/C b.1.0N/C c.3.0N/C d.12N/C

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Final answer:

The magnitude of the electric field which is 12.0 N/C at 301 cm, will be 3.0 N/C 6.00 m away from the charge, as it follows the inverse-square law.

Step-by-step explanation:

The student's question asks about the electric field at different distances from a charged object. Specifically, they want to know the value of the electric field 6.00 m away, given that at 301 cm, the field's magnitude is 12.0 N/C. The electric field due to a point charge decreases with the square of the distance from the charge, according to the formula E = k * q / r^2, where E is the electric field, k is Coulomb's constant, q is the charge, and r is the distance from the charge.

By applying the inverse-square law, we know that if the distance is doubled, the electric field's magnitude will be reduced by a factor of 4 (since 2 squared is 4). To find the magnitude at 6.00 m (which is twice the distance of 3.01 m), we can simply divide the original magnitude by 4, leading to an answer of 3.0 N/C.