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Tems11 [23]
3 years ago
12

A point charge of 4 µC is located at the center of a sphere with a radius of 25 cm. Find the electric flux through the surface o

f the sphere. The Coulomb constant is 8.98755 × 109 N · m2 /C 2 and the acceleration due to gravity is 9.8 m/s 2 .Answer in units of N · m2 /C.
Physics
1 answer:
givi [52]3 years ago
5 0

Answer:

Ф = 4.5176x10⁵ N . m² / C

Explanation:

In this case, we need to use two expressions in order to calculate the electric flux of the sphere.

First the Electric flux is calculated using this expression:

Ф = E * A  (1)

Where:

E: Electric field

A: Area of the sphere

To get the electric field E, we use this expression:

E = K * q / r²    (2)

If we replace (2) into (1) we have the following:

Ф = K * q * A / r²    (3)

Finally, we need to know the expression to get the area of a sphere which is the following:

A = 4πr²   (4)

Replacing into (3):}

Ф = K * Q * 4πr² / r²       discarting r²:

Ф = K * Q * 4π   (5)

Now, all we need to do is replace the given values and solve for the electric flux of the sphere:

Ф = 8.98755x10⁹ * 4x10⁻⁶ * 4 * π

<h2>Ф = 4.5176x10⁵ N . m² / C</h2>

Hope this helps

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Answer:

1. Reflection

2. travel from one medium to another

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1. When a wave strikes a solid barrier, it bounces back in the same medium. This wave behavior of bouncing back is known as reflection. Its like a basketball hitting a backboard. The ball bounces back at the same angle as it was incident. ∠i = ∠r

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\frac{sin(i)}{sin(r)} = \frac{\mu_2}{\mu_1}

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LUCKY_DIMON [66]

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Where θ₂ is the angle of the transmitted ray θ_p is the angle of the reflected polarized ray

We replace

     n1 sin θ_p = n2 sin (90 - θ_p)

Let's use the trigonometry relationship

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