A) \[\frac{\lambda }{2\pi {{a}^{2}}{{\varepsilon }_{0}}}\]
B) \[\frac{\lambda }{4{{\pi }^{2}}{{\varepsilon }_{0}}a}\]
C) \[\frac{\lambda }{2{{\pi }^{2}}{{\varepsilon }_{0}}a}\]
D) zero
Correct Answer: C
Solution :
[c] Considering symmetric elements each of length \[dl\] at A and B, we note that electric fields perpendicular to PO are cancelled and those along PO are added. The electric field due to an element of length \[dl(=ad\theta )\] along PO. |
\[dE=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{dq}{{{a}^{2}}}\cos \theta \] |
\[(\because dl=ad\theta )\] |
\[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\lambda dl}{{{a}^{2}}}\cos \theta \] |
\[=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{\lambda (ad\theta )}{{{a}^{2}}}\cos \theta \] |
Net electric field at O |
\[E=\int_{-\pi /2}^{\pi /2}{dE=2\int_{O}^{\pi /2}{\frac{1}{4\pi {{\varepsilon }_{0}}}}}\frac{\lambda a\cos \theta \,d\theta }{{{a}^{2}}}\] |
\[=2\cdot \frac{1}{4\pi {{\varepsilon }_{O}}}\frac{\lambda }{a}[\sin \theta ]_{o}^{\pi /2}\] |
\[=2\cdot \frac{1}{4\pi {{\varepsilon }_{O}}}\cdot \frac{\lambda }{a}\cdot 1=\frac{\lambda }{2\pi {{\varepsilon }_{o}}a}\] |
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