Question

Rod of length l has a total charge q distributed uniformly along its length it is bent in the shape of semicircle find the magnitude of electric field at the centre of curvature of semicircle

Answer 1

By conservation of length, we can say

l=pi*r.

=>r=l/pi

Now, at angle theta from the axis of symmetry of semicircle, take a differential length dl on the semicircle.

=>dl=r*d(theta)

=>dq=(q/l)*dl

You can easily calculate field due to dl as,

dE=k*dq/r^2

This dE will have two components out of which the one away from the semicircle will be net dE.

Now integrate dE from theta=-pi/2 to pi/2

You will easily get your answer

Answer 2

The magnitude of electric field at the centre of curvature of semi-circle is, $E=\frac{\theta}{2 \varepsilon_{0} L^{2}}$.

Explanation:

As we know that,

                            $\lambda=\mathrm{Q} / \mathrm{L}$

Where,        λ is nothing but the charge per unit length  

                   L is length of rod

                  Q is l charge  

                  $\mathrm{dq}_{1} \text { for a length i.e } \mathrm{d} =\lambda^{\times} \mathrm{d}l$

Center electric field due to charge = $k .^{\frac{d q}{r^{2}}}$

The horizontal Electric field components complement one another.

All that remains are vertical components.  

Net Electrical field vertically

                     $d_{E}=2 E \cos \theta$

                     $d_{E}=\mathrm{k} \frac{d q}{r^{2}} \times \cos \theta$

                     $d_{E}=\frac{2 k \cos \theta}{r^{2}} \lambda \times \mathrm{d} 1$

But d,               $\theta=\frac{d l}{r}=d l=r d \theta$

                      $d_{E}=\frac{2 k \lambda}{r^{2}} \cos \theta \times r d \theta$

                       $d_{E}=\frac{2 k \lambda}{r} \cos \theta d \theta$

Integration,

                          $E=\int_{0}^{\frac{\pi}{2}} \frac{2 k \lambda}{r} \cos \theta d \theta$

                          $E=\int_{0}^{\frac{\pi}{2}} \frac{2 k \lambda}{r} \sin \theta$

                          $E=\frac{2 k \lambda 1}{r}$

                          $E=\frac{2 k \theta}{L r}$

                    $\text { But } \mathrm{L}=\pi r$

                            $r=\frac{L}{\pi}$

                           $E=\frac{2 k \theta}{L \frac{L}{\pi}}$

                           $E=\frac{2 k \pi \theta}{L^{2}}$

                           $E=\frac{\pi \theta}{L^{2}} \frac{2}{4 \pi \varepsilon_{0}}$

                          $E=\frac{\theta}{2 \varepsilon_{0} L^{2}}$

Thus, the magnitude of electric field is, $E=\frac{\theta}{2 \varepsilon_{0} L^{2}}$ .