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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.

Arun
25763 Points
one year 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.