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What magnitude charge creates a 4.70 N/C electric field at a point 3.70 m away?

User Zeeple
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1 Answer

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Final answer:

To calculate the magnitude of the charge that creates a 4.70 N/C electric field at a point 3.70 m away, we can use Coulomb's law. The magnitude of the charge is approximately 2.58 * 10^(-6) C.

Step-by-step explanation:

To calculate the magnitude of the charge that creates a 4.70 N/C electric field at a point 3.70 m away, we can use Coulomb's law. Coulomb's law states that the electric field is equal to the force exerted by the charge, divided by the distance squared. The equation is:

E = k * (q / r^2)

where E is the electric field, q is the charge, r is the distance, and k is the electrostatic constant.

In this case, we are given the electric field (4.70 N/C) and the distance (3.70 m). Plugging these values into the equation, we can solve for the magnitude of the charge:

4.70 N/C = k * (q / (3.70 m)^2)

Simplifying the equation, we have:

q = (E * r^2) / k

q = (4.70 N/C * (3.70 m)^2) / k

Using the given values of the electrostatic constant and performing the calculation, we find that the magnitude of the charge is approximately 2.58 * 10^(-6) C.

User Hardik Mer
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