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A +14 nC charge is located at the origin.

A)What is the strength of the electric field at the position (x,y)=(5.0cm,0cm)?

User Utukku
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Final answer:

The electric field strength at the position (x,y) = (5.0 cm, 0 cm) due to a +14 nC charge located at the origin can be calculated using Coulomb's law.

Step-by-step explanation:

The strength of the electric field at the position (x,y) = (5.0 cm, 0 cm) due to a +14 nC charge located at the origin can be calculated using Coulomb's law. The formula to calculate the electric field is given by:

E = k * (Q / r²)

where E is the electric field strength, k is the electrostatic constant (9 x 10^9 Nm²/C²), Q is the charge (in this case +14 nC = 14 x 10^-9 C), and r is the distance between the charges (in this case 5.0 cm = 0.05 m).

Plugging in the values, we get:

E = (9 x 10^9 Nm²/C²) * (14 x 10^-9 C) / (0.05 m)²

Simplifying this gives us:

E = (9 x 10^9 Nm²/C²) * (14 x 10^-9 C) / (0.0025 m²)

Calculating this gives us the strength of the electric field at the given position.

User Jeannicolas
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