A) \[\frac{\sigma }{\pi {{\varepsilon }_{0}}b}(\hat{i}+\hat{j}+\hat{k})\]
B) \[\frac{\sigma }{2\pi {{\varepsilon }_{0}}b}(\hat{i}+\hat{j}+\hat{k})\]
C) \[\frac{\sigma }{\pi {{\varepsilon }_{0}}b}(\hat{i}+\hat{j}-\hat{k})\]
D) \[\frac{\sigma }{2\pi {{\varepsilon }_{0}}b}(\hat{i}+\hat{j}-\hat{k})\]
Correct Answer: B
Solution :
Electric field at P. due to wire 3 only is \[{{E}_{3}}=\frac{\sigma }{2\pi {{\varepsilon }_{0}}{{({{b}^{2}}+{{b}^{2}})}^{1/2}}}(\hat{i}\cos {{45}^{o}}+\hat{k}\cos {{45}^{o}})\] \[=\frac{\sigma }{2\sqrt{2}\pi {{\varepsilon }_{0}}b}\times \frac{1}{\sqrt{2}}(\hat{i}+\hat{k})\] Similarly, electric field due to wires 1 and 2 \[{{E}_{1}}=\frac{\sigma }{2\sqrt{2}\pi {{\varepsilon }_{0}}b}\times \frac{1}{\sqrt{2}}(\hat{j}\,+\hat{k})\] and \[{{E}_{2}}=\frac{\sigma }{2\sqrt{2}\pi {{\varepsilon }_{0}}b}\times \frac{1}{\sqrt{2}}(\hat{i}\,+\hat{k})\] \[{{E}_{net}}={{E}_{1}}+{{E}_{2}}+{{E}_{3}}=\frac{\sigma }{2\pi {{\varepsilon }_{0}}b}(\hat{i}\,+\hat{j}+\hat{k})\]
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