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Reynold's Number

Two charges o...

Question

Two charges of $10 μ C$ and $- 90 μ C$ are separated by a distance of $24 c m$ . Electrostatic field strength from the smaller charge is zero at a distance of

12 c m

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24 c m

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36 c m

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48 c m

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Solution

The correct option is A $12 c m$
Let $x$ be the distance where $E$ is zero.
$\Rightarrow E = \frac{k (10 \times 10^{- 6})}{x^{2}} (-^x) + \frac{k (90 \times 10^{- 6})}{(x + 24)^{2}} (+^x)$
$\Rightarrow 0 = - \frac{10}{x^{2}} + \frac{90}{(x + 24)^{2}}$
$\Rightarrow \frac{1}{x^{2}} = \frac{9}{(x + 24)^{2}}$
$\Rightarrow \frac{1}{x} = \frac{3}{x + 24}$
$\Rightarrow x + 24 = 3 x$
$\Rightarrow x = 12$

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Two charges of 10μC and −90μC are separated by a distance of 24 cm. Electrostatic field strength from the smaller charge is zero at a distance of

The correct option is A 12 cmLet x be the distance where E is zero.⇒E=k(10×10−6)x2(−^x)+k(90×10−6)(x+24)2(+^x)⇒0=−10x2+90(x+24)2⇒1x2=9(x+24)2⇒1x=3x+24⇒x+24=3x⇒x=12

Two charges of $10 μ C$ and $- 90 μ C$ are separated by a distance of $24 c m$ . Electrostatic field strength from the smaller charge is zero at a distance of