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arlik [135]
3 years ago
11

What is the electric field 3.9 m from the center of the terminal of a Van de Graaff with a 6.60 mC charge, noting that the field

is equivalent to that of a point charge at the center of the terminal
Physics
1 answer:
earnstyle [38]3 years ago
3 0

Answer:

the electric field is  3.91 x 10⁶ N/C

Explanation:

Given the data in the question;

Electric field at a point due to point charge is;

E = kq/r²

where k is the constant, r is the distance from centre of terminal to point where electric field is, q is the excess charge placed on the centre of terminal of Van de Graff,a generator

Now, given that r = 3.9 m, k = 9.0 x 10⁹ Nm²/C², q = 6.60 mC = 6.60 x 10⁻³ C

so we substitute into the formula

E = [(9.0 x 10⁹ Nm²/C²)( 6.60 x 10⁻³ C)] / ( 3.9 )²

E = 59400000 / 15.21

E = 3.91 x 10⁶ N/C

Therefore, the electric field is  3.91 x 10⁶ N/C

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Answer:

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Explanation:

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A real image is four times as far from a lens as is theobject.
saveliy_v [14]

Answer:

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Explanation:

The lens equation states that:

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In this problem, the image is 4 times as far from the lens as is the object: this means that

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If we substitute this into the lens equation and we rearrange it, we get

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