Physics, asked by amaythegenius, 8 months ago

Figure shows an electric quadrupole,
with quadrupole moment (Q=2qt”)
The electric field at a distance from
its centre at the axis of the
quadrupole is given by​

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Answers

Answered by topwriters
0

V∝ 1/r³ for a dipole and  V∝ 1/r for a monopole.

Explanation:

Four charges of same magnitude are placed at points X, Y, Y, and Z respectively. A point is located at P, which is r distance away from point Y. Given that the system of charges forms an electric quadrupole.

So the system of the electric quadrupole has three charges as follows:

Charge +q placed at point X

Charge −2q placed at point Y

Charge +q placed at point Z

XY = YZ = a

YP = r

PX = r+a

PZ = r-a

Electrostatic potential caused by the system of three charges at point P is given by,

 V = 1/4π∈₀ [ q/XP  - 2q/YP + q/ZP]

    =  1/4π∈₀ [ 1/r+a  - 2q/r + q/r-a]

    =  q/4π∈₀ [ r(r-a) - 2(r+a)(r-a) - r(r+a) / r(r+a)(r-a) ]

    =  q/4π∈₀ [ r²-ra - 2r² + 2a² + r² + ra] / [r(r² - a²)]

    =  q/4π∈₀ [ 2a² / r(r² - a²) ]

    =  2qa² /4π∈₀. r³(1 - a²/r²) ]      

We know that r/a >> 1, so a/r << 1.

Hence a²/r² will be negligible.

V = 2qa² /4π∈₀r³

V∝ 1/r³ for a dipole and  V∝ 1/r for a monopole.

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