A) \[12\times {{10}^{4}}\]
B) \[24\times {{10}^{4}}\]
C) \[36\times {{10}^{4}}\]
D) \[48\times {{10}^{4}}\]
Correct Answer: C
Solution :
\[E=\frac{1}{4\pi {{\varepsilon }_{0}}}.\left[ \frac{5\times {{10}^{-9}}}{{{(1\times {{10}^{-2}})}^{2}}}-\frac{5\times {{10}^{-9}}}{{{(2\times {{10}^{-2}})}^{2}}}+\frac{5\times {{10}^{-9}}}{{{(4\times {{10}^{-2}})}^{2}}} \right.\]\[\left. -\frac{(5\times {{10}^{-9}})}{{{(8\times {{10}^{-2}})}^{2}}}+..... \right]\] \[\Rightarrow E=\frac{9\times {{10}^{9}}\times 5\times {{10}^{-9}}}{{{10}^{-4}}}\left[ 1-\frac{1}{{{(2)}^{2}}}+\frac{1}{{{(4)}^{2}}}-\frac{1}{{{(8)}^{2}}}+... \right]\] \[\Rightarrow E=45\times {{10}^{4}}\left[ 1+\frac{1}{{{(4)}^{2}}}+\frac{1}{{{(16)}^{2}}}+... \right]\]\[-45\times {{10}^{4}}\left[ \frac{1}{{{(2)}^{2}}}+\frac{1}{{{(8)}^{2}}}+\frac{1}{{{(32)}^{2}}}+... \right]\] \[\Rightarrow E=45\times {{10}^{4}}\left[ \frac{1}{1-\frac{1}{16}} \right]-\frac{45\times 10{{\,}^{4}}}{(2){{\,}^{2}}}\left[ 1+\frac{1}{{{4}^{2}}}+\frac{1}{{{(16)}^{2}}}+.. \right]\] \[E=\text{ 48}\times \text{1}{{0}^{\text{4}}}\text{ 12}\times \text{1}{{0}^{\text{4}}}=\text{ 36}\times \text{1}{{0}^{\text{4}}}N/C\]
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