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Along the x-axis, three charges $\large\frac{q}{2},$$-q$ and $\large\frac{q}{2}$ are placed at $x=0,x=a$ and $x=2a$ respectively. The resultant electric potential at a point P located at a distance r from the charge -q( a << r) is ($ \varepsilon _0 $ is the permittivity of free space)

\[\begin {array} {1 1} (a)\;\frac{qa}{4 \pi \varepsilon_0 r^2} & \quad (b)\;\frac{qa^2}{4 \pi \varepsilon_0 r^3} \\ (c)\;\frac{q\bigg(\frac{a^2}{4}\bigg)}{4 \pi \varepsilon_0 r^3} & \quad (d)\;\frac{q}{4 \pi \varepsilon_0 r} \end {array}\]

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$(b)\;\frac{qa^2}{4 \pi \varepsilon_0 r^3} $
answered Nov 7, 2013 by pady_1
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