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

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(B) The electric potential $V$ at a point due to a system of charges is the algebraic sum of the potentials due to individual charges.The positions of

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

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(B) The electric potential $V$ at a point due to a system of charges is the algebraic sum of the potentials due to individual charges.The positions of the charges are $x_1 = 0$,$x_2 = a$,and $x_3 = 2a$. The point $P$ is at $x = a+r$.The distances of point $P$ from the charges are:$r_1 = (a+r) - 0 = r+a$$r_2 = (a+r) - a = r$$r_3 = (a+r) - 2a = r-a$The total potential $V_P$ is:$V_P = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q/2}{r+a} - \frac{q}{r} + \frac{q/2}{r-a} \right]$$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{1}{2(r+a)} - \frac{1}{r} + \frac{1}{2(r-a)} \right]$$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{r(r-a) - 2(r^2-a^2) + r(r+a)}{2r(r^2-a^2)} \right]$$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{r^2 - ar - 2r^2 + 2a^2 + r^2 + ar}{2r(r^2-a^2)} \right]$$V_P = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{2a^2}{2r(r^2-a^2)} \right] = \frac{q a^2}{4 \pi \varepsilon_0 r(r^2-a^2)}$Since $a \ll r$,we have $r^2 - a^2 \approx r^2$.Therefore,$V_P = \frac{q a^2}{4 \pi \varepsilon_0 r^3}$.

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

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