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A charge of 0.80 μC is located at the origin (0.00,0.00) cm. What is the magnitude of the electric field at location (40.0,20.0) cm?

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Answer:

The magnitude of the electric field at the given location (40.0, 20.0) cm from a charge of 0.80 µC is calculated to be approximately 35960 N/C using the formula E = k * |q| / r^2, where k is Coulomb's constant.

Step-by-step explanation:

To calculate the magnitude of the electric field at a certain point due to a point charge, one can use Coulomb's Law. The equation for the electric field E due to a point charge q at a distance r is given by:

E = k * |q| / r2

Where:

  • k is the Coulomb's constant (8.99 × 109 Nm2/C2),
  • q is the charge,
  • and r is the distance from the charge to the point where the field is being measured.

In the given problem, the charge is 0.80 µC (which is 0.80 × 10-6 C), and the location is (40.0, 20.0) cm from the origin. First, convert the distance to meters: the distance r is √(0.42 + 0.22) meters. Then, apply the formula to find the electric field's magnitude.

Calculating r:

r = √(0.42 + 0.22)
= √(0.16 + 0.04)
= √0.20
= 0.447 m

Calculating E:

E = (8.99 × 109 Nm2/C2) * (0.80 × 10-6 C) / (0.447 m)2
= (8.99 × 109 * 0.80 × 10-6) / 0.20
= 35960 N/C

Therefore, the magnitude of the electric field at the location (40.0, 20.0) cm is approximately 35960 N/C.

User Dario Dias
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