Three charged particles having charges q, -2q | vedclass

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Question

(B) The electric potential $V$ at a point $P(r, 0)$ due to a system of charges is the algebraic sum of the potentials due to individual charges.$V = V

Vedclass · Accepted Answer

$\frac{1}{4 \pi \varepsilon_0} \frac{2 q a^2}{r^3}$

Vedclass · Answer

(B) The electric potential $V$ at a point $P(r, 0)$ due to a system of charges is the algebraic sum of the potentials due to individual charges.$V = V_1 + V_2 + V_3$$V = \frac{1}{4 \pi \varepsilon_0} \left[ \frac{q}{r+a} + \frac{-2q}{r} + \frac{q}{r-a} \right]$$V = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{1}{r+a} + \frac{1}{r-a} - \frac{2}{r} \right]$$V = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{(r-a) + (r+a)}{r^2 - a^2} - \frac{2}{r} \right]$$V = \frac{q}{4 \pi \varepsilon_0} \left[ \frac{2r}{r^2 - a^2} - \frac{2}{r} \right]$$V = \frac{2q}{4 \pi \varepsilon_0} \left[ \frac{r^2 - (r^2 - a^2)}{r(r^2 - a^2)} \right]$$V = \frac{2q}{4 \pi \varepsilon_0} \left[ \frac{a^2}{r(r^2 - a^2)} \right]$Since $r \gg a$,we can approximate $r^2 - a^2 \approx r^2$.$V \approx \frac{2q}{4 \pi \varepsilon_0} \frac{a^2}{r^3} = \frac{1}{4 \pi \varepsilon_0} \frac{2 q a^2}{r^3}$.

Three charged particles having charges $q, -2q$,and $q$ are placed in a line at points $(-a, 0), (0, 0)$,and $(a, 0)$ respectively. The expression for electric potential at $P(r, 0)$ for $r \gg a$ is ...............

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