3 years ago

Which statement accurately describes the outer planets?

Physics

2 answers:

tiny-mole [99]3 years ago

6 0

The outer planets have a high gravity due to their large size

True [87]3 years ago

5 0

Answer:

Its D on Edge 2020 just took the quiz

Explanation:

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2 years ago

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A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge

Kisachek [45]

Answer:

$\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

Explanation:

The electric field created by an infinitely long wire can be found by Gauss' Law.

$\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r$

For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.

$\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$