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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2019
  • Hard
  • MCQ

Four point charges \(-q, +q, +q\) and \(-q\) are placed on \(y\) axis at \(y = -2d\), \(y = -d, y = +d\) and \(y = +2d\), respectively. The magnitude of the electric field \(E\) at a point on the \(x -\) axis at \(x = D\), with \(D > > d\), will vary as

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Answer & Solution

Correct Answer

(D) \(E \propto \frac{1}{D^4}\)

Step-by-step Solution

Detailed explanation

Electric field at \(\mathrm{p}=2 \mathrm{E}_{1} \cos \theta_{1}-2 \mathrm{E}_{2} \cos \theta_{2}\)

\(=\frac{2 \mathrm{Kq}}{\left(\mathrm{d}^{2}+\mathrm{D}^{2}\right)} \times \frac{\mathrm{D}}{\left(\mathrm{d}^{2}+\mathrm{D}^{2}\right)^{1 / 2}}-\frac{2 \mathrm{Kq}}{\left[(2 \mathrm{d})^{2}+\mathrm{D}^{2}\right]} \times \frac{\mathrm{D}}{\left[(2 \mathrm{d})^{2}+\mathrm{D}^{2}\right]^{1 / 2}}\)

\(=2 \mathrm{KqD}\left[\left(\mathrm{d}^{2}+\mathrm{D}^{2}\right)^{-3 / 2}-\left(4 \mathrm{d}^{2}+\mathrm{D}^{2}\right)^{-3 / 2}\right]\)

\(=\frac{2 \mathrm{KqD}}{\mathrm{D}^{3}}\left[\left(1+\frac{\mathrm{d}^{2}}{\mathrm{D}^{2}}\right)^{-3 / 2}-\left(1+\frac{4 \mathrm{d}^{2}}{\mathrm{D}^{2}}\right)^{-3 / 2}\right]\)

Applying binomial approximation \(\because \mathrm{d}<<\mathrm{D}\)

\(=\frac{2 \mathrm{KqD}}{\mathrm{D}^{3}}\left[1-\frac{3}{2} \frac{\mathrm{d}^{2}}{\mathrm{D}^{2}}-\left(1-\frac{3 \times 4 \mathrm{d}^{2}}{2 \mathrm{D}^{2}}\right)\right]\)

\(=\frac{2 \mathrm{KqD}}{\mathrm{D}^{3}}\left[\frac{12}{2} \frac{\mathrm{d}^{2}}{\mathrm{D}^{2}}-\frac{3}{2} \frac{\mathrm{d}^{2}}{\mathrm{D}^{2}}\right]\)

\(=\frac{9 \mathrm{kqd}^{2}}{\mathrm{D}^{4}}\)

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