Find the value of $b$ where the field intensity is maximum.

\frac{1}{{(\frac{1 + a}{a})}^{2 / 3} - 1}

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\frac{1}{{(\frac{a}{1 + a})}^{2 / 3} - 1}

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\frac{1}{{(\frac{1 + a}{a})}^{1 / 3} - 1}

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\frac{1}{{(\frac{a}{1 + a})}^{1 / 3} - 1}

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Solution

The correct option is A $\frac{1}{{(\frac{1 + a}{a})}^{2 / 3} - 1}$
Field at point $x$ unit distance from $Q_{2}$ to its right is
$F = - \frac{1}{4 π ϵ} \frac{Q_{1}}{(1 + x)^{2}} + \frac{1}{4 π ϵ} \frac{Q_{2}}{x^{2}}$
$\frac{d F}{d x} = 0 \Rightarrow 2 \frac{Q_{1}}{(1 + x)^{3}} - 2 \frac{Q_{2}}{x^{3}} = 0$
$\frac{(1 + a)^{2}}{a^{2}} = \frac{Q_{1}}{Q_{2}} = \frac{(1 + x)^{3}}{x^{3}}$
$∵$ Field is $z e r o$ at point $a$ ,
$\frac{1}{4 π ϵ} \frac{Q_{1}}{(1 + a)^{2}} = \frac{1}{4 π ϵ} \frac{Q_{2}}{a^{2}}$
$∣ ∣ ∣ \frac{Q_{1}}{Q_{2}} ∣ ∣ ∣ = \frac{(1 + a)^{2}}{a^{2}})$
Therefore ,
$(\frac{1 + a}{a})^{2 / 3} = \frac{1 + x}{x}$
$x = \frac{1}{(\frac{1 + a}{a})^{2 / 3} - 1}$

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Find the value of b where the field intensity is maximum.

Find the value of $b$ where the field intensity is maximum.