Question

Calculate the electric flux through ring shown in figure is:

• A. $\frac{q}{2{\epsilon }_0}[1+\frac{L}{\sqrt{R^2+L^2}}]$
• B. $\frac{q}{2{\epsilon }_0}[1-\frac{L}{\sqrt{R^2+L^2}}]$
• C. $\frac{q}{{\epsilon }_0}[1-\frac{L}{\sqrt{R^2+L^2}}]$
• D. Zero

Answer 1

The correct option is A $\frac{q}{2{\epsilon }_0}[1+\frac{L}{\sqrt{R^2+L^2}}]$

Electric flux through the elemental ring is $d\varphi =Ed\cos \theta$

$=\frac{kq}{L^2+R^2}\frac{RdR}{{(l^2+R^2)}^{3/2}}$

Total flux the ring

$Q=\int d\varphi =\frac{dl}{2{ϵ}_0}{\int }_0^R\frac{RdR}{{(l^2+R^2)}^{3/2}}$

$\varphi =\frac{ql}{2{ϵ}_0}{[-\frac{1}{\sqrt{l^2+R^2}}]}_0^R$

$=\frac{q}{2{ϵ}_0}[1+\frac{L}{\sqrt{l^2+R^2}}]$