On a surface I have two point charges Qa and Qb, 1C and 4C respectivly and are placed at A(-3,0) B(3,0). Find all the points (locus) (except those that are too far away) where V=0.?

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Answer 1 · 2018-08-04T12:16:26

Technically, no such locus exists.

Assuming V stands for the total electric potential, then there is no answer to this question unless one of the charges is negative.

The total electric potential V at any point (x,y) is:

$V = V_{a} + V_{b}$ $V = V_a + V_b$

$= \frac{Q_{a}}{r_{a}} + \frac{Q_{b}}{r_{b}}$ $" "= Q_a/r_a + Q_b/r_b$

where $r_{a}$ $r_a$ and $r_{b}$ $r_b$ are distance of $Q_{a}$ $Q_a$ and $Q_{b}$ $Q_b$ from the point (x,y) respectively.

Since $Q_{a}$ $Q_a$ and $Q_{b}$ $Q_b$ are both positive, V can never be zero unless $r_{a}$ $r_a$ & $r_{b} \to \infty$ $r_b ->oo$

Hence, there is no solution to your question unless one of the charges is negative. Then

$V = \frac{Q_{a}}{r_{a}} - \frac{Q_{b}}{r_{b}} = 0$ $V = Q_a/r_a - Q_b/r_b =0$

or $- \frac{Q_{a}}{r_{a}} + \frac{Q_{b}}{r_{b}} = 0$ $- Q_a/r_a + Q_b/r_b =0$

Then you can solve the equation above for the locus by setting

$r_{a} = \sqrt{{(x + 3)}^{2} + y^{2}}$ $r_a= sqrt((x+3)^2+y^2)$
$r_{b} = \sqrt{{(x - 3)}^{2} + y^{2}}$ $r_b=sqrt((x-3)^2+y^2)$