WBJEE · Physics · Electrostatics
Three point charges \(q,-2 q\) and \(q\) are placed along \(x\) axis at \(x=-a, 0\) and a respectively. As a \(\rightarrow 0\) and \(q \rightarrow \infty\) while \(q a^2=Q\) remains finite, the clectric field at a point \(P\), at a distance \(x(x \gg a)\) from \(x=0\) is \(E=\frac{\alpha Q}{4 \pi \in_0 x \beta} \hat{i}\). Then
- A \(\alpha=\beta\)
- B \(\alpha=2 \beta\)
- C \(\alpha=\frac{2}{3} \beta\)
- D \(2 \alpha=2 \beta\)
Answer & Solution
Correct Answer
(D) \(2 \alpha=2 \beta\)
Step-by-step Solution
Detailed explanation
Hint : \(E=\frac{2 k q a}{x^3}\left[\frac{1}{\left(1-\frac{a}{2 x}\right)^3}-\frac{1}{\left(1+\frac{a}{2 x}\right)^3}\right]\)…
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