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What magnitude charge creates a 8.60 n/c electric field at a point 4.60 m away?

User AI Snoek
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1 Answer

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Final answer:

The magnitude of the charge needed to create an electric field of 8.60 n/C at a distance of 4.60 m is 2.07 × 10^-11 Coulombs. This is calculated using Coulomb's law, which relates the electric field to the charge creating it and the distance from the charge.

Step-by-step explanation:

To determine the magnitude of the charge that creates an electric field of 8.60 n/C at a distance of 4.60 m, we use Coulomb's law of electric fields:

E = k · q / r^2,

where E is the electric field, k is Coulomb's constant (8.99 × 10^9 N·m^2/C^2), q is the charge, and r is the distance from the charge.

First, solve for q:

q = E · r^2 / k,

then plug in the given values:

q = (8.60 n/C) · (4.60 m)^2 / (8.99 × 10^9 N·m^2/C^2).

Converting n/C to C:
8.60 n/C = 8.60 × 10^-9 C/C,

Perform the calculation:

q = (8.60 × 10^-9 C/C) · (4.60 m)^2 / (8.99 × 10^9 N·m^2/C^2),

q = 2.07 × 10^-11 C.

The magnitude of the charge needed to create this electric field is 2.07 × 10^-11 Coulombs.

User Cyrille Pontvieux
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