A) \[12\times {{10}^{9}}q\text{ }N/C\]
B) Zero
C) 6 ´ 109q N/C
D) 4 ´ 109q N/C
Correct Answer: A
Solution :
Net field at origin \[E=\frac{q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{1}{{{1}^{2}}}+\frac{1}{{{2}^{2}}}+\frac{1}{{{4}^{2}}}+....\infty \right]\] \[=\frac{q}{4\pi {{\varepsilon }_{0}}}\left[ 1+\frac{1}{4}+\frac{1}{16}+.....\infty \right]\] \[=\frac{q}{4\pi {{\varepsilon }_{0}}}\left[ \frac{1}{1-\frac{1}{4}} \right]=12\times {{10}^{9}}q\,N/C\]
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