The electric intensity due to a dipole of len | vedclass

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Question

(A) Given: Dipole length $2l = 10 \, cm = 0.1 \, m$,so $l = 0.05 \, m$. Charge $q = 500 \, \mu C = 5 	imes 10^{-4} \, C$. Distance from the center of

Vedclass · Accepted Answer

$6.25 	imes 10^7 \, N/C$

Vedclass · Answer

(A) Given: Dipole length $2l = 10 \, cm = 0.1 \, m$,so $l = 0.05 \, m$. Charge $q = 500 \, \mu C = 5 	imes 10^{-4} \, C$. Distance from the center of the dipole $r = 20 \, cm + 5 \, cm = 25 \, cm = 0.25 \, m$.Dipole moment $p = q 	imes 2l = (5 	imes 10^{-4} \, C) 	imes (0.1 \, m) = 5 	imes 10^{-5} \, C \cdot m$.The electric field on the axial line of a dipole is given by $E = \frac{1}{4\pi\epsilon_0} \cdot \frac{2pr}{(r^2 - l^2)^2}$.Substituting the values: $E = (9 	imes 10^9) 	imes \frac{2 	imes (5 	imes 10^{-5}) 	imes 0.25}{((0.25)^2 - (0.05)^2)^2}$.$E = (9 	imes 10^9) 	imes \frac{2.5 	imes 10^{-5}}{(0.0625 - 0.0025)^2} = (9 	imes 10^9) 	imes \frac{2.5 	imes 10^{-5}}{(0.06)^2}$.$E = \frac{22.5 	imes 10^4}{0.0036} = 6.25 	imes 10^7 \, N/C$.

The electric intensity due to a dipole of length $10 \, cm$ and having a charge of $500 \, \mu C$,at a point on the axis at a distance $20 \, cm$ from one of the charges in air,is

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