- Home
- Standard 12
- Physics
A wire of length $L\, (=20\, cm)$, is bent into a semicircular arc. If the two equal halves of the arc were each to be uniformly charged with charges $ \pm Q\,,\,\left[ {\left| Q \right| = {{10}^3}{\varepsilon _0}} \right]$ Coulomb where $\varepsilon _0$ is the permittivity (in $SI\, units$) of free space] the net electric field at the centre $O$ of the semicircular arc would be
$\left( {50 \times {{10}^3}\,N/C} \right)\hat j$
$\left( {50 \times {{10}^3}\,N/C} \right)\hat i$
$\left( {25 \times {{10}^3}\,N/C} \right)\hat j$
$\left( {25 \times {{10}^3}\,N/C} \right)\hat i$
Solution
Given: Length of wire $L=20 \,\mathrm{cm}$ charge $Q=10^{3} \varepsilon_{0}$
We know, electric field at the centre of the semicircular arc
${E=\frac{2 K \lambda}{r}}$
${\text{ or, }}\quad E = \frac{{2K\left( {\frac{{2Q}}{{\pi r}}} \right)}}{r}\left[ {As\,\lambda = \frac{{2Q}}{{\pi r}}} \right]$
$=\frac{4 K Q}{\pi r^{2}}=\frac{4 K Q \pi^{2}}{\pi L^{2}}=\frac{4 \pi K Q}{L^{2}}=25 \times 10^{3} \,N / C \hat{i}$
