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Grade: 12
        
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.
4 months ago

Answers : (1)

Arun
24739 Points
							
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.
 
 
4 months ago
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