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Two point – like charges Q₁ and Q₂ are positioned at point 1 and 2 respectively. The field intensity to the right of the charge Q₂ on the line that passes through the two charges varies according to a law that is represented schematically in the figure. The field intensity is assumed to be positive if its direction coincides with the positive direction on the x- axis. The distance between the charges is l, then which of the following is true

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The ratio of the absolute value of the charges $|\frac{Q_{1}}{Q_{2}}| = {(\frac{a + ℓ}{a})}^{2}$

The value of b, where the field intensity is maximum is $b = \frac{ℓ}{{(\frac{ℓ + a}{a})}^{\frac{2}{3}} - 1}$

Q₁ is positive charge while Q₂ is negative charge

The value of b, where the field intensity is maximum is $b = \frac{ℓ}{{(\frac{ℓ + a}{2 a})}^{2 / 3} - 1}$

answer is A, B, D.

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Detailed Solution

At null point, $\frac{Q_{1}}{x} = \frac{Q_{2}}{(L - x)}$ $\frac{Q_{1}}{(L + a_{2})} + \frac{Q_{2}}{a_{2}} = 0 i.e. |Q_{n}| < |Q_{1}|$

charge are opposite as null point is external

$\begin{array}{l} E = \frac{Q_{1}}{(L + a + x)^{2}} - \frac{Q_{2}}{(a + x)^{2}} \\ for E_{max} \Rightarrow \frac{dE}{dx} = 0 \end{array}$

On solving we get

$x = \frac{ℓ}{{(\frac{ℓ + a}{2 a})}^{2 / 3} - 1}$

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