A) \[\frac{\sigma }{\pi {{\varepsilon }_{0}}b}(\widehat{\mathbf{i}}+\widehat{\mathbf{j}}+\widehat{\mathbf{k}})\]
B) \[\frac{\sigma }{2\pi {{\varepsilon }_{0}}b}(\widehat{\mathbf{i}}+\widehat{\mathbf{j}}+\widehat{\mathbf{k}})\]
C) \[\frac{\sigma }{\pi {{\varepsilon }_{0}}b}(\widehat{\mathbf{i}}+\widehat{\mathbf{j}}-\widehat{\mathbf{k}})\]
D) \[\frac{\sigma }{2\pi {{\varepsilon }_{0}}b}(\widehat{\mathbf{i}}+\widehat{\mathbf{j}}-\widehat{\mathbf{k}})\]
Correct Answer: B
Solution :
Electric field at\[P\], due to wire 3 only is \[{{\mathbf{E}}_{\mathbf{3}}}=\frac{\sigma }{2\pi {{\varepsilon }_{0}}{{({{b}^{2}}+{{b}^{2}})}^{1/2}}}(\widehat{\mathbf{i}}\cos {{45}^{o}}+\widehat{\mathbf{k}}\cos {{45}^{o}})\] Similarly, electric field due to wires 1 and 2 \[{{\mathbf{E}}_{\mathbf{1}}}=\frac{\sigma }{2\sqrt{2}\pi {{\varepsilon }_{0}}b}\times \frac{1}{\sqrt{2}}(\widehat{\mathbf{j}}+\widehat{\mathbf{k}})\] and \[{{\mathbf{E}}_{\mathbf{2}}}=\frac{\sigma }{2\sqrt{2}\pi {{\varepsilon }_{0}}b}\times \frac{1}{\sqrt{2}}(\widehat{\mathbf{i}}+\widehat{\mathbf{j}})\] \[{{\mathbf{E}}_{net}}={{\mathbf{E}}_{\mathbf{1}}}+{{\mathbf{E}}_{\mathbf{2}}}+{{\mathbf{E}}_{\mathbf{3}}}=\frac{\sigma }{2\pi {{\varepsilon }_{0}}b}(\widehat{\mathbf{i}}+\widehat{\mathbf{j}}+\widehat{\mathbf{k}})\]
You need to login to perform this action.
You will be redirected in
3 sec
