A charge of $+ 2 μ C$ is placed at x=0 and a charge of $- 32 μ C$ at x=60 cm. Athird charge -Q be placed on the x-axis such that it experiences no force. The distance of the point from $+ 2 μ C$ is (in cm)

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Solution

The correct option is B 20
$_{A} = K \frac{(2 r c) (θ)}{d^{2} c m^{2}}^j = \frac{2 k θ \times 10^{- 6}}{d^{2} \times 10^{- 4}}^i$
$_{B} = - K \frac{(32 r c) θ^i}{(60 + d)^{2} c m^{2}} = \frac{- 32 l θ}{(60 + d)^{2} \times 10^{- 4}}^i$
${\to F}_{n e t} = {\to F}_{A} + {\to F}_{B} = 0$
$\Rightarrow 2 \frac{(K θ) 10^{- 6}}{d^{2} θ 10^{- 4}} - \frac{32 K θ \times 10^{- 6}}{(60 + d)^{2} \times 10^{- 4}} = 0$
$\Rightarrow \frac{1}{d^{2}} - \frac{16}{(60 + d)^{2}} = 0$
$\Rightarrow \frac{1}{d} = \frac{4}{60 + d}$
$\Rightarrow d + 60 = 4 d$
$\Rightarrow d = 20$ cm
Option B is correct.

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A charge of +2μC is placed at x=0 and a charge of −32μC at x=60 cm. Athird charge -Q be placed on the x-axis such that it experiences no force. The distance of the point from +2μC is (in cm)

The correct option is B 20→FA=K(2rc)(θ)d2cm2^j=2kθ×10−6d2×10−4^i→FB=−K(32rc)θ^i(60+d)2cm2=−32lθ(60+d)2×10−4^i→Fnet=→FA+→FB=0⇒2(Kθ)10−6d2θ10−4−32Kθ×10−6(60+d)2×10−4=0⇒1d2−16(60+d)2=0⇒1d=460+d⇒d+60=4d⇒d=20cmOption B is correct.

A charge of $+ 2 μ C$ is placed at x=0 and a charge of $- 32 μ C$ at x=60 cm. Athird charge -Q be placed on the x-axis such that it experiences no force. The distance of the point from $+ 2 μ C$ is (in cm)