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BartSMP [9]
3 years ago
15

A certain electric dipole consists of charges q and –q separated by distance d, oriented along the x-axis as shown in the figure

. find an expression for the magnitude of the electric field e of the dipole at a point far away in the y-direction, a distance r away from the midpoint of the dipole. assume that r is much greater than
d. enter your answer in terms of q, d, r, and ε0.
Physics
1 answer:
Anna71 [15]3 years ago
7 0

Electric field due to a point charge is given by formula

E = \frac{kq}{r^2 + \frac{d^2}{4}}

now due to negative charge also the magnitude of electric field will be same

only difference is the direction of field due to negative charge is radially inwards

now we can say that net field due to these two charges is given as

E = 2E_0cos\theta

E = 2\frac{kq}{r^2 + \frac{d^2}{4}}*\frac{d/2}{\sqrt{r^2 + \frac{d^2}{4}}}

E = \frac{kqd}{(r^2 + \frac{d^2}{4})^1.5}

now it is given that distance r is very large than "d" so we can say

E = \frac{kqd}{r^3}

<em>so above is the electric field due to dipole</em>

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An ice skater of mass m = 60 kg coasts at a speed of v = 0.8 m/s past a pole. At the distance of closest approach, her center of
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Answer:

1.78 rad/s

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v = Velocity = 0.8 m/s

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r_2 = New distance between center of mass of person and pole = 0.46 m

I = Moment of inertia

Angular speed is given by

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The angular speed is 1.78 rad/s

In this system the angular momentum is conserved

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7 0
3 years ago
The magnitude of the Normal Force on a
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Answer:

5. Is greater than mg, always

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y-axis:

N*cos β - m*g = m*ay   where ay is the vertical acceleration. If the block starts falling down, ay will be negative. If the block starts sliding up, ay will be positive. If the block does not move up nor down, ay=0.

Solving for N:

N = \frac{m*g + m*ay}{cos \beta }

If ay is positive or zero, N will be greater than mg. If ay is negative, N will be less than mg.

If the block is sliding along a horizontal circular path (not up, nor down), ay = 0, so N will always be greater than mg.

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Answer:

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