Question

A thin glass rod is bent into a semicircle of radius r. A charge +Q is uniformly distributed along the upper half and a charge –Q is uniformly distributed along the lower half, as shown in the figure. The electric field E at, P (the centre of the semicircle, is)

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

Answer 1

The correct option is A $\frac{Q}{{\pi }^2{\epsilon }_0{r}^2}$

Let us consider a small element dq on the upper half of the semicircular ring at an angle $\theta$ with the horizontal.
The electric field due to this element will be dE.
Now consider a similar element in the lower part of the ring and the electric field due to this element will also be dE as shownOn taking the component of this electric field we will get dE $\cos \theta$as horizontal component and dE$\sin \theta$ as vetrical component.
Only vertical components will add up and horzontal component will get cancelled.
$⟹d{E}_{net}=2dE\sin \theta$........(i)
The electric field dE due to the small element dq will be
$dE=\frac{kdq}{r}$...(ii)
Let dq be at an angle of d$\theta$ at the center
$⟹dq=\frac{Q}{\frac{\pi r}{2}}d\theta$.......(iii)
Putting (iii) in (ii) we get
$dE=\frac{2Qd\theta }{4\pi {ϵ}_0{r}^{(}2)}$....(iv)
putting (iv) in (i) and integrating we get
$\int d{E}_{net}={\int }_0^{\frac{\pi }{2}}\frac{Q\sin \theta d\theta }{(\pi {)}^2{r}^2{ϵ}_0}$
$⟹E_{net}=\frac{Q}{(\pi {)}^2{ϵ}_0{r}^2}$