Two charges $2.0 \times 10^{- 6} C$ and $10 \times 10^{- 6} C$ are placed at a separation of $10 c m$ . Where should a third charge be placed such that it experiences no net force due to these charges?

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Solution

If both are positive charge then the third charge should be placed on the line joining two charge.
$F_{1} = F_{2}$
$\frac{2 q}{r^{2}} = \frac{1 q}{(10 - r)^{2}}$
$r^{2} = 2 (10 - r)^{2}$
$r^{2} = 200 + 2 r^{2} - 40 r$
$r^{2} - 40 r + 200 = 0$
$r = \frac{[40 \pm \sqrt{(1600 - 800)}]}{2}$
$r = \frac{[40 - \sqrt{(1600 - 800)}]}{2}$ Since the point should be in between
$r = 20 - 10 \sqrt{2}$
$r = 5.86 c m$

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Two charges 2.0×10−6C and 10×10−6C are placed at a separation of 10 cm. Where should a third charge be placed such that it experiences no net force due to these charges?

If both are positive charge then the third charge should be placed on the line joining two charge.F1=F22qr2=1q(10−r)2r2=2(10−r)2r2=200+2r2−40rr2−40r+200=0r=[40±√(1600−800)]2r=[40−√(1600−800)]2 Since the point should be in betweenr=20−10√2r=5.86cm

Two charges $2.0 \times 10^{- 6} C$ and $10 \times 10^{- 6} C$ are placed at a separation of $10 c m$ . Where should a third charge be placed such that it experiences no net force due to these charges?