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agasfer [191]
3 years ago
9

Help me please i need this to passsss

Physics
1 answer:
mamaluj [8]3 years ago
6 0

Answer:   D

Explanation:     : ⊇

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Given that,

Rank in order from largest to smallest the magnitude of the electric field at block dot.

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Electric field is proportional to the charge divided by square of distance.

In mathematically,

E\propto\dfrac{q}{r^2}

Where, q = charge

r = distance

If the charge is greater then electric field will be greater.

If the distance is greater then electric field will be smaller.

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According to figure,

(I). The electric field at black dot due to positive charge point q to left. the distance is r.

The electric field will be

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(II). The electric field at black dot due to positive charge point 2q to left. The distance is 2r.

Then, the electric field will be

E=\dfrac{k2q}{(2r)^2}

E=\dfrac{kq}{2r^2}

The electric field will be smallest.

(III).  The electric field at black dot due to positive charge point 2q to left. The distance is r.

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(IV). The electric field at black dot due to positive charge point q to left. The distance is 2r.

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E=\dfrac{kq}{(2r)^2}

E=\dfrac{kq}{4r^2}

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E_{3}>E_{1}>E_{2}>E_{4}

Hence, The ranking will be 3, 1, 2, 4.

(D) is correct option.

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