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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
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Question : 22 of 37
Marks: +1, -0
Figure shows a charge array known as an electric quadrupole. For a point on the axis of quadrupole, obtain the dependence of potential on r for r/a >> 1, and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).
Solution:  
VPV_P = VPA+VPB+VPC+VPDV_{PA} + V_{PB} + V_{PC} + V_{PD}
or VPV_P = 14πε0\frac{1}{4\pi\varepsilon_0} [+qr+a−qr−qr+qr−a]\left[ \frac{+q}{r+a} - \frac{q}{r} - \frac{q}{r} + \frac{q}{r-a} \right]
or VPV_P =
q4πε0[r(r−a)−2(r2−a2)+r(r+a)r(r2−a2)]\frac{q}{4\pi\varepsilon_0} \left[ \frac{r(r-a)-2(r^2-a^2)+r(r+a)}{r(r^2-a^2)} \right]
or VPV_P =
q4πε\frac{q}{4\pi\varepsilon_{}} [r2−ra−2r2+a2+r2+rar(r2−a2)]\left[ \frac{r^2 - ra - 2r^2 + a^2 + r^2 + ra}{r(r^2 - a^2)} \right]
or VPV_P = 14πε0\frac{1}{4\pi\varepsilon_0} q⋅2a2r(r2−a2)\frac{q \cdot 2a^2}{r(r^2 - a^2)} = 14πε0p⋅ar(r2−a2)\frac{1}{4\pi\varepsilon_0} \frac{p \cdot a}{r(r^2 - a^2)}
For ra\frac{r}{a} ≫ or r ≫ a
VPV_P ~ 14πε0\frac{1}{4\pi\varepsilon_0} par3\frac{pa}{r^3} or VPV_P α 1r3\frac{1}{r^3}
However, electric potential at any point on axis of electric dipole is
V = 14πε0pr2\frac{1}{4\pi\varepsilon_0} \frac{p}{r^2} or V α 1r2\frac{1}{r^2} and due to point charge is
V = 14πε0qr\frac{1}{4\pi\varepsilon_0} \frac{q}{r} or V α 1r\frac{1}{r}
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