Question

Find the electric field at a given point ‘O’ due to a semi-circular charged ring having linear charge density $\lambda$.

• A. $\frac{\lambda }{4\pi {\epsilon }_{0}r}$
• B. $\frac{\lambda }{2{\epsilon }_{0}r}$
• C. $\frac{\lambda }{2\pi {\epsilon }_{0}r}$
• D. $\frac{\lambda }{4\pi {\epsilon }_0{r}^2}$

Answer 1

The correct option is C

$\frac{\lambda }{2\pi {\epsilon }_{0}r}$

We know that the electric field due to a circular arc at a point on its Centre

E = $\frac{\lambda }{2\pi {\epsilon }_{0}r}\sin \frac{\alpha }{2}$

where α is the angle subtended at the Centre. Here the angle subtended at the Centre is $\pi$

[It’s ok if you don’t remember this, I recommend you to watch the derivation video once again]

So

E = $\frac{\lambda }{2\pi {\epsilon }_{0}r}\sin \frac{\pi }{2}$

= $\frac{\lambda }{2\pi {\epsilon }_{0}r}$