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An infinite number of electric charges each equal to $5\, nC$ (magnitude) are placed along $X$-axis at $x = 1$ $cm$, $x = 2$ $cm$ , $x = 4$ $cm$ $x = 8$ $cm$ ………. and so on. In the setup if the consecutive charges have opposite sign, then the electric field in Newton/Coulomb at $x = 0$ is $\left( {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,N - {m^2}/{c^2}} \right)$
$12 \times {10^4}$
$24 \times {10^4}$
$36 \times {10^4}$
$48 \times {10^4}$
Solution
(c) $E = \frac{1}{{4\pi {\varepsilon _0}}}.\left[ {\frac{{5 \times {{10}^{ – 9}}}}{{{{(1 \times {{10}^{ – 2}})}^2}}} – \frac{{5 \times {{10}^{ – 9}}}}{{{{(2 \times {{10}^{ – 2}})}^2}}} + \frac{{5 \times {{10}^{ – 9}}}}{{{{(4 \times {{10}^{ – 2}})}^2}}}} \right.$ $\left. { – \frac{{(5 \times {{10}^{ – 9}})}}{{{{(8 \times {{10}^{ – 2}})}^2}}} + …..} \right]$
$ \Rightarrow E = \frac{{9 \times {{10}^9} \times 5 \times {{10}^{ – 9}}}}{{{{10}^{ – 4}}}}\left[ {1 – \frac{1}{{{{(2)}^2}}} + \frac{1}{{{{(4)}^2}}} – \frac{1}{{{{(8)}^2}}} + …} \right]$
$ \Rightarrow E = 45 \times {10^4}\left[ {1 + \frac{1}{{{{(4)}^2}}} + \frac{1}{{{{(16)}^2}}} + …} \right]$
$ – 45 \times {10^4}\left[ {\frac{1}{{{{(2)}^2}}} + \frac{1}{{{{(8)}^2}}} + \frac{1}{{{{(32)}^2}}} + …} \right]$
$ \Rightarrow E = 45 \times {10^4}\left[ {\frac{1}{{1 – \frac{1}{{16}}}}} \right] – \frac{{45 \times 10{\,^4}}}{{(2){\,^2}}}\left[ {1 + \frac{1}{{{4^2}}} + \frac{1}{{{{(16)}^2}}} + ..} \right]$
$E = 48 \times 10^4 -12 \times 10^4 = 36 \times 10^4\, N/C$
