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Aneli [31]
2 years ago
8

Find the magnitude of the electric field at points along the axis of a dipole (along the same line that contains +Q and −Q) for

r≫ℓ, where r is the distance from a point to the center of the dipole and ℓ is the distance between +Q and −Q.
Physics
1 answer:
Vaselesa [24]2 years ago
5 0

Answer:

E=\frac{2K\cdot P}{r^3}

Explanation:

We are given that a dipole consist of two charge Q and -Q and charge separated by l.Let a charge +1 C is placed at  point P at distance r from the centre of dipole.

We have to find the magnitude of the electric field at point along the axis of dipole .

We know that Electric field=\frac{Force}{unit \;positive\;Charge}

Electric filed due to positive charge Q

E_1=\frac{kQ}{(r+\frac{\rho}{2})^2} {from A to P}

Electric field due to negative charge -Q

E_2=\frac{KQ}{(r-\frac{\rho}{2})^2} ( Along PB)

Net electric field E=E_2-E_1

E=\frac{kQ}{(r-\frac{\rho}{2})^2}-\frac{KQ}{(r+\frac{\rho}{2})^2}

E=KQ\frac{r^2+\rho^2+r\rho-r^2-\rho^2+r\rho}{(r^2-\frac{\rho^2}{4})^2}

E=\frac{2 KQ r\rho}{(r^2-\frac{\rho^2}{4})^2}

We are given that r >> L then

E=\frac{2KQ r\rho}{r^4}=\frac{2kQ \rho}{r^3}

E=\frac{K\cdot 2Q\rho}{r^3}

E=\frac{2K\cdot P}{r^3}

Where, P=Diple = Distance between two charges \times any charge

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Using equilibrium of force in vertical direction for left spring

k_{L} x_{L} = (0.5) W\\(59) x_{L} = (0.5) (13.72)\\x_{L} = 0.116 m

Using equilibrium of force in vertical direction for right spring

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