Question

A uniform long wire of length $1m$ carrying a charge of $1.1C$ is bent into a circular loop, with two ends leaving a $10cm$ gap between them. The electric field at the centre of the circle, then must be:

• A. $51.43\times {10}^{9}V/m$
• B. $42.43\times {10}^{9}V/m$
• C. $32.314\times {10}^{9}V/m$
• D. $24.35\times {10}^{9}V/m$

Answer 1

The correct option is C $32.314\times {10}^{9}V/m$

Given,

$2\pi r=1+0.1=1.1m\Rightarrow r=\frac{1.1}{2\pi }m$

Total charge of ring,$Q=1.1C$

$\therefore$ linear charge density ${\lambda }_{+}=\frac{1.1}{1}=1.1m^{-1},{\lambda }_{-}=-1.1m^{-1}$

Hence field at O ,$E_O=$Field due to ring of ${\lambda }_{+}+$ Field due to arc AB of ${\lambda }_{-}$

$=0+\frac{{\lambda }_{-}}{4\pi {\epsilon }_0{r}^2}{\int }_{{\theta }_1}^{{\theta }_2}rd\theta =\frac{{\lambda }_{-}}{4\pi {\epsilon }_{0}r}.({\theta }_2-{\theta }_1)=\frac{{\lambda }_{-}}{4\pi {\epsilon }_0}\frac{\Delta \theta }{r}$

$\therefore E=-\frac{1.1}{4\pi {\epsilon }_0}.\frac{r\Delta \theta }{r^2}=-\frac{1.1}{4\pi {\epsilon }_0}.\frac{0.1}{r^2}$

$\therefore |E_O|=1.1\times 9\times {10}^9\times \frac{0.1}{1.21}\times 4\times \frac{22}{7}\times \frac{22}{7}$ as, $r=\frac{1.1}{2\pi }$

$=\frac{11\times 9\times {10}^9\times 4\times 22\times 22}{121\times 7\times 7}=32.314\times {10}^{9}V{m}^{-1}$