Question

Figure shows a plastic rod having a uniformly distributed charge $Q$. The rod has been bent in a ${120}^o$ circular arc of radius $r$. In terms of $Q$ and $r$, what is the electric field $\overrightarrow{E}$ due to the rod at point $P$?

• A. $\overrightarrow{E}=\frac{3\sqrt{3}\lambda }{8{\pi }^2{ϵ}_0{r}^2}\hat{i}$ where $\lambda =\frac{2Q}{3\pi r}$
• B. $\overrightarrow{E}=\frac{3\sqrt{3}\lambda }{8{\pi }^2{ϵ}_0{r}^2}\hat{i}$ where $\lambda =\frac{3Q}{2\pi r}$
• C. $\overrightarrow{E}=\frac{3\sqrt{3}\lambda }{8{\pi }^2{ϵ}_0{r}^2}\hat{j}$ where $\lambda =\frac{2Q}{3\pi r}$
• D. $\overrightarrow{E}=\frac{3\sqrt{3}\lambda }{8{\pi }^2{ϵ}_0{r}^2}\hat{j}$ where $\lambda =\frac{3Q}{2\pi r}$

Answer 1

The correct option is A $\overrightarrow{E}=\frac{3\sqrt{3}\lambda }{8{\pi }^2{ϵ}_0{r}^2}\hat{i}$ where $\lambda =\frac{2Q}{3\pi r}$

$E=\int dE\cos \theta$

$=\frac{C\lambda }{4\pi {ϵ}_0{r}^2}\hat{i}$

$=\frac{3\sqrt{3}\lambda }{8\pi {ϵ}_0{r}^2}\hat{i}$

where $\lambda =\frac{2Q}{3\pi r}$