Along the $x$-axis, three charges $\frac{q}{2}, - q\ and\frac{q}{2}$ 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

Question

Along the $x$-axis, three charges $\frac{q}{2}, - q\ and\frac{q}{2}$ 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)

Dr. Abubacker · Accepted Answer

(b) $As\ r \gg a,so\ r > 2a$
So, potential at point p
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$V = \frac{1}{4\pi\varepsilon_{0}}\left\lbrack \frac{\frac{q}{2}}{r + a} - \frac{q}{r} + \frac{\frac{q}{2}}{r - a} \right\rbrack$
$= \frac{1}{4\pi\varepsilon_{0}}\frac{q}{2}\left\lbrack \frac{1}{r + a} - \frac{2}{r} + \frac{1}{r - a} \right\rbrack$
$= \frac{q}{8\pi\varepsilon_{0}}\left\lbrack \frac{r(r - a) - 2\left( r^{2} - a^{2} \right) + r(r + a)}{r\left( r^{2} - a^{2} \right)} \right\rbrack$
$= \frac{q}{8\pi\varepsilon_{0}}.\frac{2a^{2}}{r\left( r^{2} - a^{2} \right)}$
$= \frac{qa^{2}}{4\pi\varepsilon_{0}r^{3}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ (as\ r \gg a)$

Q. Along the $x$-axis, three charges $\frac{q}{2}, - q\ and\frac{q}{2}$ 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)

Explanation

Reviewed by Dr. Abubacker

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