A) \[12\times {{10}^{9}}q\,N/C\]
B) zero
C) \[6\times {{10}^{9}}q\,N/C\]
D) \[4\times {{10}^{9}}q\,N/C\]
Correct Answer: A
Solution :
If there are n point charges \[{{q}_{1}},{{q}_{2}},.....{{q}_{n}},\] then each of them will produce the same intensity at any point which it would have produced in the absence of other point charges. Hence, total intensity will be vector sum of \[\overrightarrow{{{E}_{1}}},\overrightarrow{{{E}_{2}}}.....,\overrightarrow{{{E}_{n}}}\] produced at a point. \[\Sigma E=\frac{q}{4\pi {{\varepsilon }_{0}}r_{1}^{2}}+\frac{q}{4\pi {{\varepsilon }_{0}}r_{2}^{2}}+.....+\frac{q}{4\pi {{\varepsilon }_{0}}r_{n}^{2}}\] \[\therefore \] \[\vec{E}=\frac{q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{1}{{{1}^{2}}}+\frac{1}{{{2}^{2}}}+\frac{1}{{{4}^{2}}}+\frac{1}{{{8}^{2}}}+.... \right]\] The given series is a geometric progression. Hence, \[sum\,(S)=\frac{a}{1-r}\] where a is first term of series and r the common difference. \[r=\frac{1}{4},\,\,\,a=1\] \[\therefore \] \[S=\frac{1}{1-\frac{1}{4}}=\frac{4}{3}\] \[\therefore \] Total intensity \[=\frac{q\times 4}{4\pi {{\varepsilon }_{0}}\times 3}\] \[=\frac{9\times {{10}^{9}}\times 4q}{3}\] \[=12\times {{10}^{9}}q\,N/C\]
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