Question

A wire, of length $L(=20cm)$, is bent into a semi-circular arc. If the two equal halves, of the arc, were each to be uniformly charged with charges $\pm Q,[|Q|={10}^3{ϵ}_0$ Coulomb where ${ϵ}_0$ is the permittivity (in SI units) of free space] the net electric field at the centre O of the semi-circular arc would be :

• A. $(25\times {10}^{3}N/C)\hat{i}$
• B. $(25\times {10}^{3}N/C)\hat{j}$
• C. $(50\times {10}^{3}N/C)\hat{j}$
• D. $(50\times {10}^{3}N/C)\hat{i}$

Answer 1

Answer: (A)

$(25\times {10}^{3}N/C)\hat{i}$

Electric field due to quarter ring is
(fig.)
Where $K=9\times {10}^9=\frac{1}{4\pi {\epsilon }_0}$ and $\lambda =$linear charge density
$=$Charge per unit length
By super-position theorem, electric field at centre due to combination of above 2 quarters is given as $\overrightarrow{E_{net}}=\overrightarrow{E_1}+\overrightarrow{E_2}$
$E_{net}=\frac{2K\lambda }{r}\hat{i}$
$=\frac{2K\left(\frac{2Q}{\pi r}\right)}{r}=\frac{4KQ}{\pi r^2}=\frac{4KQ{\pi }^2}{\pi L^2}$
$\therefore E_{net}=\frac{4\pi KQ}{L^2}=25\times {10}^{3}N/C\hat{i}$
(Thus option A)