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Question

(A) The electric field at point $P$ at distance $x$ due to the three charges is:$E = \frac{1}{4 \pi \epsilon_0} \left[ \frac{q}{(x-a)^2} - \frac{2q}{x

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$\alpha = \beta$

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(A) The electric field at point $P$ at distance $x$ due to the three charges is:$E = \frac{1}{4 \pi \epsilon_0} \left[ \frac{q}{(x-a)^2} - \frac{2q}{x^2} + \frac{q}{(x+a)^2} \right]$$E = \frac{q}{4 \pi \epsilon_0} \left[ \frac{(x+a)^2 - 2(x^2-a^2) + (x-a)^2}{x^2(x^2-a^2)} \right]$$E = \frac{q}{4 \pi \epsilon_0} \left[ \frac{x^2 + 2ax + a^2 - 2x^2 + 2a^2 + x^2 - 2ax + a^2}{x^2(x^2-a^2)} \right]$$E = \frac{q}{4 \pi \epsilon_0} \left[ \frac{4a^2}{x^2(x^2-a^2)} \right]$Since $x \gg a$, we can approximate $x^2 - a^2 \approx x^2$:$E = \frac{q}{4 \pi \epsilon_0} \left[ \frac{4a^2}{x^2(x^2)} \right] = \frac{4qa^2}{4 \pi \epsilon_0 x^4}$Given $qa^2 = Q$, we have $E = \frac{4Q}{4 \pi \epsilon_0 x^4}$.Comparing this with $E = \frac{\alpha Q}{4 \pi \epsilon_0 x^\beta}$, we get $\alpha = 4$ and $\beta = 4$.Therefore, $\alpha = \beta$.

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