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Q.

Three charges +5q, +Q & -2q are kept along a straight line in the same order such that +5q & -2q charges are at a distance of $\frac{2 r}{3} & \frac{r}{3}$ from the charge Q respectively. If the net force on the charges -2q is zero, Then +Q is

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a

-3q

b

3q

c

$- \frac{5}{9} q$

d

$+ \frac{5}{9} q$

answer is A.

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Detailed Solution

The given situation is shown below

$F_{n e t} = 0$
$F (5 q, - 2 q) = F (Q, - 2 q)$
$\frac{K 5 q (- 2 q)}{r^{2}} = \frac{- k q (2 q)}{r^{2}}$
$10 q = 2 \times 9 \times Q$
$Q = \frac{5}{9} q$

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