A) \[\frac{q}{2\pi {{\varepsilon }_{0}}}\]
B) \[\frac{q}{3\pi {{\varepsilon }_{0}}}\]
C) \[\frac{q}{4\pi {{\varepsilon }_{0}}}\]
D) \[\frac{q}{8\pi {{\varepsilon }_{0}}}\]
Correct Answer: B
Solution :
As the charges are placed along the same straight line, the electric field at \[\operatorname{x}= 0\] will be directed along x-axis and its magnitude is given by \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}\left[ \frac{q}{{{1}^{2}}}+\frac{q}{{{2}^{2}}}+\frac{q}{{{4}^{2}}}+\frac{q}{{{8}^{2}}}+......\,\,\infty \right]\] \[=\,\,\frac{q}{4\pi {{\varepsilon }_{0}}}\left[ 1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}\,\,+\,\,......\,\,\infty \right]\] \[=\,\,\frac{q}{4\pi {{\varepsilon }_{0}}}\,\,\left( \frac{1}{1-\frac{1}{4}} \right)=\frac{q}{3\pi {{\varepsilon }_{0}}}\]You need to login to perform this action.
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