Two charges of +25 times 10^{-9} , C and -25 | vedclass

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(A) Given: Charge $q = 25 	imes 10^{-9} \, C$,distance between charges $2l = 6 \, m$,so $l = 3 \, m$. The distance from the center is $r = 4 \, m$.Th

Vedclass · Accepted Answer

$\frac{1000}{49}$

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(A) Given: Charge $q = 25 	imes 10^{-9} \, C$,distance between charges $2l = 6 \, m$,so $l = 3 \, m$. The distance from the center is $r = 4 \, m$.The electric field on the axial line is given by $E_{	ext{axial}} = \frac{1}{4\pi\epsilon_0} \frac{2pr}{(r^2 - l^2)^2}$,where $p = q(2l)$.Substituting values: $E_{	ext{axial}} = \frac{k \cdot 2(q \cdot 2l)r}{(r^2 - l^2)^2} = \frac{k \cdot 2(25 	imes 10^{-9} 	imes 6) 	imes 4}{(4^2 - 3^2)^2} = \frac{k \cdot 1200 	imes 10^{-9}}{(16 - 9)^2} = \frac{k \cdot 1200 	imes 10^{-9}}{49}$.The electric field on the equatorial line is given by $E_{	ext{eq}} = \frac{1}{4\pi\epsilon_0} \frac{p}{(r^2 + l^2)^{3/2}}$.Substituting values: $E_{	ext{eq}} = \frac{k \cdot (25 	imes 10^{-9} 	imes 6)}{(4^2 + 3^2)^{3/2}} = \frac{k \cdot 150 	imes 10^{-9}}{(16 + 9)^{3/2}} = \frac{k \cdot 150 	imes 10^{-9}}{125}$.Ratio $\frac{E_{	ext{axial}}}{E_{	ext{eq}}} = \frac{1200 	imes 10^{-9} / 49}{150 	imes 10^{-9} / 125} = \frac{1200}{49} 	imes \frac{125}{150} = 8 	imes \frac{125}{49} = \frac{1000}{49}$.

Two charges of $+25 \times 10^{-9} \, C$ and $-25 \times 10^{-9} \, C$ are placed $6 \, m$ apart. Find the ratio of the electric field intensity at a point $4 \, m$ from the center of the electric dipole on the $(i)$ axial line and $(ii)$ equatorial line.

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