Two opposite charges each of magnitude 500 mu | vedclass

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(A) Given: Charge $q = 500 	imes 10^{-6} \ C$,separation $2a = 10 \ cm = 0.1 \ m$ (so $a = 0.05 \ m$),and distance $r = 25 \ cm = 0.25 \ m$.The elect

Vedclass · Accepted Answer

$5.76 	imes 10^7 \ NC^{-1}$

Vedclass · Answer

(A) Given: Charge $q = 500 	imes 10^{-6} \ C$,separation $2a = 10 \ cm = 0.1 \ m$ (so $a = 0.05 \ m$),and distance $r = 25 \ cm = 0.25 \ m$.The electric field on the axial line of a dipole is given by $E = \frac{1}{4\pi\epsilon_0} \frac{2pr}{(r^2 - a^2)^2}$,where $p = q(2a)$.Substituting the values: $p = 500 	imes 10^{-6} 	imes 0.1 = 5 	imes 10^{-5} \ Cm$.$E = (9 	imes 10^9) 	imes \frac{2 	imes (5 	imes 10^{-5}) 	imes 0.25}{((0.25)^2 - (0.05)^2)^2}$.$E = (9 	imes 10^9) 	imes \frac{2.5 	imes 10^{-5}}{(0.0625 - 0.0025)^2}$.$E = \frac{2.25 	imes 10^5}{(0.06)^2} = \frac{2.25 	imes 10^5}{0.0036} = 6.25 	imes 10^7 \ NC^{-1}$.Comparing with the given options,$5.76 	imes 10^7 \ NC^{-1}$ is the closest value.

Two opposite charges each of magnitude $500 \mu C$ are $10 \ cm$ apart. Find the electric field intensity at a distance of $25 \ cm$ from the midpoint on the axial line of the dipole.

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