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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 $x=a+r$ if ( a << r) is : $( \varepsilon_0$ is the permittivity of free space)

\[\begin {array} {1 1} (a)\;\frac{qa}{4 \pi \varepsilon_0r^2} & \quad (b)\;\frac{qa^2}{4 \pi \varepsilon _0 r^3} \\ (c)\;\frac{q \bigg(\frac{a^2}{4}\bigg)}{4 \pi \varepsilon_0 r^3} & \quad (d)\;\frac{q}{4 \pi \varepsilon_0 r} \end {array}\]
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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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