Determine the electric field intensity at point \(P\) due to quadruple distribution shown in figure for \(r…
- 0
- \(k q a^{2} / r^{4}\)
- \(6 \mathrm{kq} a^{2} / \mathrm{r}^{4}\)
- \(6 k q a^{2} / r^{2}\)
Solution
\(\begin{array}{l}
\overrightarrow{\mathrm{E}}_{1}=\frac{\mathrm{q}}{4 \pi \varepsilon_{0}(\mathrm{r}-\mathrm{a})^{2}} \hat{\mathrm{i}} \\
\overrightarrow{\mathrm{E}}_{2}=\frac{2 \mathrm{q}}{4 \pi \varepsilon_{0} \mathrm{r}^{2}} \hat{\mathrm{i}}, \quad \overrightarrow{\mathrm{E}}_{3}=\frac{\mathrm{q}}{4 \pi \varepsilon_{0}(\mathrm{r}+\mathrm{a})^{2}} \hat{\mathrm{i}}
\end{array}\)
Now,
\(\begin{array}{l}
\vec{E}=\vec{E}_{1}+\vec{E}_{2}+\vec{E}_{3} \\
\qquad=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{(r-a)^{2}}-\frac{2}{r^{2}}+\frac{1}{(r+a)^{2}}\right]^{2} \hat{\mathrm{i}} \\
E=\frac{q}{4 \pi \varepsilon_{0} r^{2}}\left[\left\{1-\left(\frac{a}{r}\right)\right\}^{-2}-2+\left\{1+\left(\frac{a}{r}\right)\right\}^{-2}\right]
\end{array}\)
If \(r > > a\), we can use binomial approximation :
\(\left.(1+\alpha)^{\mathrm{n}}=1+\mathrm{n} \alpha+\frac{\mathrm{n}(\mathrm{n}-1)}{2 !} \alpha^{2}+\ldots . . \text { for } \alpha< < 1\right)\)
Therefore,
\(\begin{array}{l}
E=\frac{q}{4 \pi \varepsilon_{0} r^{2}}\left[\begin{array}{c}
\left\{\left(1-2\left(-\frac{a}{r}\right)\right)+\frac{-2(-2-1)}{2}\left(\frac{-a}{r}\right)^{2}\right\} \\
-2+\left\{1-2 \frac{a}{r}+\frac{-2(-2-1)}{2}\left(\frac{a}{r}\right)^{2}\right\}
\end{array}\right] \\
=\frac{6 a^{2} q}{4 \pi \varepsilon_{0} r^{4}}=\frac{6 \mathrm{kqa}^{2}}{r^{4}}
\end{array}\)
Asked in: JEE Mains - Electrostatics - Chapter Test