The electric intensity due to a dipole of length 10 cm and
having a charge of 500 μC, at a point on the axis at a distance
20 cm from one of the charges in air, is
(a) 6.25 × 10⁷ N/C (b) 9.28 × 10⁷ N/C
(c) 13.1 × 10¹¹ N/C (d) 20.5 × 10⁷ N/C
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Answer:
N/C
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ans - (a) 6.25 × 10^7
explanation : Given : Length of the dipole (2l) =10cm = 0.1m or l = 0.05 m
Charge on the dipole (q) = 500 µC = 500 ×10–6 C and distance of the point on the axis from the mid-point of the dipole (r) = 20 + 5 = 25 cm = 0.25 m.
We know that the electric field intensity due to dipole on the given point
(E) = (1/ 4π€•)×[2(q.2l)r] /
(r^2 - l^2)2
=9 × 10^ 9 ×[2(500×10^-6 ×0.1) × 0.25] / [( 0.25 ^2 )+ (0.05)^2]^2
=(225 ×10^3)/(3.6 ×10^3)
=6.25 ×10^7 N/C
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