Question
A charged wire is bent in the form of a semicircular arc of radius $$a.$$ If charge per unit length is $$\lambda $$ coulomb/metre, the electric field at the centre $$O$$ is
A.
$$\frac{\lambda }{{2\pi {a^2}{\varepsilon _0}}}$$
B.
$$\frac{\lambda }{{4{\pi ^2}{\varepsilon _0}a}}$$
C.
$$\frac{\lambda }{{2\pi {\varepsilon _0}a}}$$
D.
zero
Answer :
$$\frac{\lambda }{{2\pi {\varepsilon _0}a}}$$
Solution :
Considering symmetric elements each of length $$dl$$ at $$A$$ and $$B,$$ we know that electric fields perpendicular to $$PO$$ are cancelled and those along $$PO$$ are added. The electric field due to an element of length $$dl\left( { = ad\theta } \right)$$ along $$PO.$$
$$\eqalign{ & dE = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{d\theta }}{{{a^2}}}\cos \theta \cr & = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\lambda dl}}{{{a^2}}}\cos \theta \cr & = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\lambda \left( {a\,d\theta } \right)}}{{{a^2}}}\cos \theta \,\,\,\left( {\because dl = a\,d\theta } \right) \cr} $$
Net electric field at $$O$$
$$\eqalign{ & E = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} d E = 2\int_0^{\frac{\pi }{2}} {\frac{1}{{4\pi {\varepsilon _0}}}} \frac{{\lambda a\cos \theta \,d\theta }}{{{a^2}}} \cr & = 2 \cdot \frac{1}{{4\pi {\varepsilon _0}}}\frac{\lambda }{a}\left[ {\sin \theta } \right]_0^{\frac{\pi }{2}} \cr & = 2 \cdot \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{\lambda }{a} \cdot 1 = \frac{\lambda }{{2\pi {\varepsilon _0}a}} \cr} $$
Considering symmetric elements each of length $$dl$$ at $$A$$ and $$B,$$ we know that electric fields perpendicular to $$PO$$ are cancelled and those along $$PO$$ are added. The electric field due to an element of length $$dl\left( { = ad\theta } \right)$$ along $$PO.$$
$$\eqalign{ & dE = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{d\theta }}{{{a^2}}}\cos \theta \cr & = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\lambda dl}}{{{a^2}}}\cos \theta \cr & = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\lambda \left( {a\,d\theta } \right)}}{{{a^2}}}\cos \theta \,\,\,\left( {\because dl = a\,d\theta } \right) \cr} $$
Net electric field at $$O$$
$$\eqalign{ & E = \int_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} d E = 2\int_0^{\frac{\pi }{2}} {\frac{1}{{4\pi {\varepsilon _0}}}} \frac{{\lambda a\cos \theta \,d\theta }}{{{a^2}}} \cr & = 2 \cdot \frac{1}{{4\pi {\varepsilon _0}}}\frac{\lambda }{a}\left[ {\sin \theta } \right]_0^{\frac{\pi }{2}} \cr & = 2 \cdot \frac{1}{{4\pi {\varepsilon _0}}} \cdot \frac{\lambda }{a} \cdot 1 = \frac{\lambda }{{2\pi {\varepsilon _0}a}} \cr} $$
