An infinite number of electric charges each e | vedclass

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Question

(c) $E = \frac{1}{{4\pi {\varepsilon _0}}}.\left[ {\frac{{5 	imes {{10}^{ - 9}}}}{{{{(1 	imes {{10}^{ - 2}})}^2}}} - \frac{{5 	imes {{10}^{ - 9}}}}

Vedclass · Accepted Answer

$36 	imes {10^4}$

Vedclass · Answer

(c) $E = \frac{1}{{4\pi {\varepsilon _0}}}.\left[ {\frac{{5 	imes {{10}^{ - 9}}}}{{{{(1 	imes {{10}^{ - 2}})}^2}}} - \frac{{5 	imes {{10}^{ - 9}}}}{{{{(2 	imes {{10}^{ - 2}})}^2}}} + \frac{{5 	imes {{10}^{ - 9}}}}{{{{(4 	imes {{10}^{ - 2}})}^2}}}} ight.$ $\left. { - \frac{{(5 	imes {{10}^{ - 9}})}}{{{{(8 	imes {{10}^{ - 2}})}^2}}} + .....} ight]$
$ \Rightarrow E = \frac{{9 	imes {{10}^9} 	imes 5 	imes {{10}^{ - 9}}}}{{{{10}^{ - 4}}}}\left[ {1 - \frac{1}{{{{(2)}^2}}} + \frac{1}{{{{(4)}^2}}} - \frac{1}{{{{(8)}^2}}} + ...} ight]$
$ \Rightarrow E = 45 	imes {10^4}\left[ {1 + \frac{1}{{{{(4)}^2}}} + \frac{1}{{{{(16)}^2}}} + ...} ight]$
$ - 45 	imes {10^4}\left[ {\frac{1}{{{{(2)}^2}}} + \frac{1}{{{{(8)}^2}}} + \frac{1}{{{{(32)}^2}}} + ...} ight]$
$ \Rightarrow E = 45 	imes {10^4}\left[ {\frac{1}{{1 - \frac{1}{{16}}}}} ight] - \frac{{45 	imes 10{\,^4}}}{{(2){\,^2}}}\left[ {1 + \frac{1}{{{4^2}}} + \frac{1}{{{{(16)}^2}}} + ..} ight]$
$E = 48 	imes 10^4 -12 	imes 10^4 = 36 	imes 10^4\, N/C$

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