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kodGreya [7K]
3 years ago
9

What is the potential energy at the origin due to an electric field of 5 x 10^6 N/C located at x=43cm,y=28cm?

Physics
1 answer:
Fudgin [204]3 years ago
3 0

Answer:

potential energy at origin is 2.57*10^{6} volt

Explanation:

given data:

electric field E = 5*10^{6} N/C

at x = 43 cm, y = 28 cm

distance btween E and origin

\Delta r = \sqrt{43^2 +28^2}

\Delta r = 51.313 cm

potential energy per unit charge \Delta V = - Edr

\Delta V = 5*10^6*51.313*10^{-2} J/C

\Delta V  =  2.57*10^{6} volt

potential energy at origin is 2.57*10^{6} volt

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The distance mirror M2 must be moved so that one wavelength has produced one more new maxima than the other wavelength is;

<u><em>L = 57.88 mm</em></u>

<u><em /></u>

We are given;

Wavelength 1; λ₁ = 589 nm = 589 × 10⁻⁹ m

Wavelength 2; λ₂ = 589.6 nm = 589.6 × 10⁻⁹ m

We are told that L₁ = L₂. Thus, we will adopt L.

Formula for the number of bright fringe shift is;

m = 2L/λ

Thus;

For Wavelength 1;

m₁ = 2L/(589 × 10⁻⁹)

For wavelength 2;

m₂ = 2L/(589.6)

Now, we are told that one wavelength must have produced one more new maxima than the other wavelength. Thus;

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Plugging in the values of m₁ and m₂ gives;

(2L/589) - (2L/589.6) = 2

divide through by 2 to get;

L[(1/589) - (1/589.6)] = 1

L(1.728 × 10⁻⁶) = 1

L = 1/(1.728 × 10⁻⁶)

L = 578790.67 nm

L = 57.88 mm

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2 years ago
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