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earnstyle [38]
4 years ago
8

The rectangular region of the xy plane shown has nonuniform surface charge density σ = σ0 (1+ y b), where σ0 is a constant. Find

the net charge.
Physics
1 answer:
sergiy2304 [10]4 years ago
8 0

Answer:

This is net charge on the surface  is  Q = σ₀ x (y + 2by²)

Explanation:

The surface charge density is defined as the amount of charge Q per unit area A

       σ = dq / dA

       dq = σ dA

Since the surface is a rectangular region we use an xy coordinate system so the area difference  

      dA = dxdy

      dq = σ dx dy

 We replace, evaluate the integral

        ∫ dq = ∫ σ₀ (1 + yb) dxdy

realizamos laintegral de dx

        Q -0 =σ₀ ∫ (1 + yb) (x-0)   dy

Where we evaluate We must recognize that the charge Q must be zero by the time X = 0 and Y = 0. At the starting point Q = 0 for x = 0

 

We perform the other integral (dy)

        Q = σ₀ x (y + 2y² b)

Evaluated between Y = 0 and Y = y

      Q = σ₀ x (y + 2by²)

This is net charge on the surface

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4 years ago
How to remember physics derivation easier? <br>​
polet [3.4K]

Answer:

hmm let's see

Explanation:

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3 years ago
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To find the final velocity of the second ball you have to use the conversation of momentum:

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Let one ball be ball 1 and the other be ball 2

m₁ = 0.17kg

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Since the mass of the balls are the same we can factor it out and get rid of the numbers below it so....

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The masses now cancel because we factored them out on both sides so if we divide mass over to another side the value will cancel out so....

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Now we want the final velocity of the second ball so we need v₂f

so...

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Plug in the numbers now:

(0.75 + 0.65) - 0.5 = v₂f

v₂f = 0.9 m/s


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