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An infinite number of electric charges each equal to 5 nano-coulomb (magnitude) are placed along $X$-axis at $x=1 \mathrm{~cm}, x=2 \mathrm{~cm}$, $x=4 \mathrm{~cm}, x=8 \mathrm{~cm} . . .$. and so on. In this set up if the consecutive charges have opposite sign, then the electric field in newton/coulomb at $x=0$ is :
$\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}^2\right)$
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$\left(\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~N}-\mathrm{m}^2 / \mathrm{C}^2\right)$
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Verified Answer
The correct answer is:
$36 \times 10^4$
Electric field intensity due to a point charge.
$E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}$
The consecutive charges are of opposite signs.
$\therefore$ Net electric field at $x=0$, is
$E=\frac{1}{4 \pi \varepsilon_0}\left[\frac{Q}{r_1^2}-\frac{Q}{r_2^2}+\frac{Q}{r_3^2}-\frac{Q}{r_4^2} \ldots . . \infty\right]$
$=\frac{Q}{4 \pi \varepsilon_0}\left[\frac{1}{r_1^2}-\frac{1}{r_2^2}+\frac{1}{r_3^2}-\frac{1}{r_4^2}+\ldots . \infty\right\rceil$
$=9 \times 10^9 \times 5 \times 10^{-9}$
$\left[\frac{1}{\left(1 \times 10^{-2}\right)^2}-\frac{1}{\left(2 \times 10^{-2}\right)^2}+\frac{1}{\left(4 \times 10^{-2}\right)^2}+\ldots \infty\right\rfloor$
$=\frac{45}{10^{-4}}\left[\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{8^2}+\ldots . \infty\right]$
$=45 \times 10^4\left[\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{8^2} \ldots \infty\right]$
$\left\lceil\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{8^2} \ldots \infty\right\rfloor=\frac{\frac{1}{1^2}}{1-\left(\frac{-1}{2}\right)^2}$
$\left[\right.$ Sum of infinite GP, $\left.\mathrm{S}_{\infty}=\frac{a}{1-r}\right]$
$=\frac{1}{1+\frac{1}{4}}=\frac{4}{5}$
$\therefore \quad E=45 \times 10^4 \times \frac{4}{5}=36 \times 10^4 \mathrm{~N} / \mathrm{C}$
$E=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{Q}{r^2}$
The consecutive charges are of opposite signs.
$\therefore$ Net electric field at $x=0$, is
$E=\frac{1}{4 \pi \varepsilon_0}\left[\frac{Q}{r_1^2}-\frac{Q}{r_2^2}+\frac{Q}{r_3^2}-\frac{Q}{r_4^2} \ldots . . \infty\right]$
$=\frac{Q}{4 \pi \varepsilon_0}\left[\frac{1}{r_1^2}-\frac{1}{r_2^2}+\frac{1}{r_3^2}-\frac{1}{r_4^2}+\ldots . \infty\right\rceil$
$=9 \times 10^9 \times 5 \times 10^{-9}$
$\left[\frac{1}{\left(1 \times 10^{-2}\right)^2}-\frac{1}{\left(2 \times 10^{-2}\right)^2}+\frac{1}{\left(4 \times 10^{-2}\right)^2}+\ldots \infty\right\rfloor$
$=\frac{45}{10^{-4}}\left[\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{8^2}+\ldots . \infty\right]$
$=45 \times 10^4\left[\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{8^2} \ldots \infty\right]$
$\left\lceil\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{4^2}-\frac{1}{8^2} \ldots \infty\right\rfloor=\frac{\frac{1}{1^2}}{1-\left(\frac{-1}{2}\right)^2}$
$\left[\right.$ Sum of infinite GP, $\left.\mathrm{S}_{\infty}=\frac{a}{1-r}\right]$
$=\frac{1}{1+\frac{1}{4}}=\frac{4}{5}$
$\therefore \quad E=45 \times 10^4 \times \frac{4}{5}=36 \times 10^4 \mathrm{~N} / \mathrm{C}$
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