Along the x-axis,three charges frac{q}{2}, -q | vedclass

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(B) The charges are located at $x=0$ $(q/2)$,$x=a$ $(-q)$,and $x=2a$ $(q/2)$.The point $P$ is at a distance $r$ from the charge $-q$ at $x=a$. Since $

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$\frac{q a^2}{4 \pi \varepsilon_0 r^3}$

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(B) The charges are located at $x=0$ $(q/2)$,$x=a$ $(-q)$,and $x=2a$ $(q/2)$.The point $P$ is at a distance $r$ from the charge $-q$ at $x=a$. Since $P$ is on the $x$-axis to the right of the charges,its coordinate is $x_P = a + r$.The distances of point $P$ from the three charges are:$1$. From $q/2$ at $x=0$: $d_1 = (a+r) - 0 = r+a$$2$. From $-q$ at $x=a$: $d_2 = (a+r) - a = r$$3$. From $q/2$ at $x=2a$: $d_3 = (a+r) - 2a = r-a$The total electric potential $V$ at point $P$ is the sum of potentials due to individual charges:$V = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q/2}{r+a} - \frac{q}{r} + \frac{q/2}{r-a} \right]$$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{1}{r+a} - \frac{2}{r} + \frac{1}{r-a} \right]$$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{r(r-a) - 2(r^2-a^2) + r(r+a)}{r(r^2-a^2)} \right]$$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{r^2 - ar - 2r^2 + 2a^2 + r^2 + ar}{r(r^2-a^2)} \right]$$V = \frac{q}{8 \pi \varepsilon_0} \left[ \frac{2a^2}{r(r^2-a^2)} \right]$Since $r \gg a$,we can approximate $r^2 - a^2 \approx r^2$:$V \approx \frac{q}{8 \pi \varepsilon_0} \cdot \frac{2a^2}{r^3} = \frac{q a^2}{4 \pi \varepsilon_0 r^3}$

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 $-q$ (where $r > a$) is ($\varepsilon_0$ is the permittivity of free space):

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