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svetoff [14.1K]
3 years ago
10

Charge of uniform density (0.30 nC/m2) is distributed over the xy plane, and charge of uniform density (−0.40 nC/m2) is distribu

ted over the yz plane. What is the magnitude of the resulting electric field at any point not in either of the two charged planes?
Physics
1 answer:
soldier1979 [14.2K]3 years ago
8 0

Answer: E = 39.54 N/C

Explanation: Electric field can be determined using surface charge density:

E = \frac{\sigma}{2\epsilon_{0}}

where:

σ is surface charge density

\epsilon_{0} is permitivitty of free space (\epsilon_{0} = 8.85.10^{-12}C^{2}/N.m^{2})

Calculating resulting electric field:

E=E_{1} - E_{2}

E = \frac{\sigma_{1}-\sigma_{2}}{2\epsilon_{0}}

E = \frac{[0.3-(-0.4)].10^{-9}}{2.8.85.10^{-12}}

E=0.03954.10^{3}

E = 39.54

The resulting Electric Field at any point is 39.54N/C.

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Given :

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3 years ago
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Answer:

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Let´s find the vertical distance traveled by Brad each day:

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R=3818Km

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