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calculate the electric field intensity due to a dipole of length 10cm and having a charge of 500 micro coulomb at a point on the axis distant 20 cm from one of the charges in air.

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7 months ago

```							Length if dipole = 10cm Distance from point of axis = 20cm Charge = 500 μC The electric field intensity due a dipole at a point on the axial line is given as - E = 2kpr/(r²-l²)2 where p is the dipole moment of dipole, r is the separation between midpoint of dipole to the observation point. 500 μC on each pole of dipole and separation between two dipoles will be 2l = 10cm Dipole moment , p = 500 × 100cm = 500 × 10^-6 C × 0.1 m = 5 × 10^-5 C.m Distance between the observation point to one charge, d = 20cm Distance between observation points to midpoint of dipole moment , r = (d + x/2) = 20cm + 5cm = 25cm = 0.25m E = 2 × 9 × 10^9 × 5 × 10^-5 × 0.25/(0.25² - 0.05²)² = 90 × 10⁴ × 0.25/(0.30)²(0.20)² = 2.25 × 10^5/(0.09 × 0.04) = 2.25/(36) × 10^9 = 225/36 × 10^7 = 6.25 × 10^7 N/C Thus, one of the charges in air, is 6.25 × 10^7 N/C.
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7 months ago
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