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DENIUS [597]
3 years ago
5

Rank in order, from largest to smallest, the magnitudes of the electric field at the black dot. A. 2, 1, 3, 4 B. 1, 4, 2, 3 C. 3

, 2, 1, 4 D. 3, 1, 2, 4

Physics
1 answer:
sweet [91]3 years ago
4 0

Given that,

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

Electric field :

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.

We need to find the electric field at black dot

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

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

The electric field will be largest.

(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.

Then, the electric field will be

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

The electric field will be very largest.

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

Then, the electric field will be

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

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

The electric field will be very smallest.

So, The electric field from largest to smallest will be

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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