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A particle has a charge of +8.00 nC and is located at the origin of an x-y grid. What is the magnitude of the electric field due to this particle at the location in space given by the coordinates (x = +3.00 cm, y = +4.00 cm)?

User Halena
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Step-by-step explanation:

To calculate the magnitude of the electric field at a point in space due to a charged particle, you can use the formula:

Electric Field (E) = (k * |q|) / r^2

Where:

- E is the electric field.

- k is the electrostatic constant (approximately 8.99 x 10^9 N·m^2/C^2).

- |q| is the magnitude of the charge (in this case, 8.00 nC, which is 8.00 x 10^-9 C).

- r is the distance between the point and the charged particle.

In this case, the point is located at (x = +3.00 cm, y = +4.00 cm), and you can calculate the distance (r) as follows:

r = √((x^2 + y^2))

r = √((0.03 m)^2 + (0.04 m)^2)

r = √((0.0009 m^2 + 0.0016 m^2))

r = √(0.0025 m^2)

r = 0.05 m

Now, you can calculate the electric field magnitude:

E = (8.99 x 10^9 N·m^2/C^2 * 8.00 x 10^-9 C) / (0.05 m)^2

E ≈ (7.192 x 10^1 N·m^2/C) / (0.0025 m^2)

E ≈ 2.8768 x 10^4 N/C

So, the magnitude of the electric field due to the +8.00 nC charge at the point (x = +3.00 cm, y = +4.00 cm) is approximately 2.8768 x 10^4 N/C.

Hopefully i helped u out! =)

User Imanexcelnoob
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