A thin glass rod is bent into a semicircle of radius R.A charge Q is uniformly distributed along upper half and a charge -Q is uniformly distributed along lower half.Calculate elecric field E at Pt. P the centre of semicircle is

Dear Student,

Please find below the solution to the asked query:

The situation is as shown in the figure below.



As shown in the figure, the upper part is charged with positive charge and the lower part with negative charge. Consider the symmetrical elements one of the positive charge and the other is of negative charge.

The electric field due to the element of the positive charge is,

$d{E}_{+}=\frac{1}{4\pi {\epsilon }_0}\frac{dQ}{R^2}$

The electric field due to the element of the negative charge is,

$d{E}_{-}=\frac{1}{4\pi {\epsilon }_0}\frac{dQ}{R^2}$

These two will be as shown in the figure, the components of these two along the horizontal axis are equal and opposite. Therefore, these two will get cancel each other.

The two components are along the same direction and equal in magnitude. Therefore, these two components will get added to give the resultant. Therefore the effective electric field is,

$$\begin{gathered} dE=d{E}_{+} \cos \theta +d{E}_{-} cos \theta =\frac{2}{4\pi {\epsilon }_0}\frac{\lambda R}{R^2} \cos \theta d\theta \\ \Rightarrow E={\int }_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^{0}dE=\frac{1}{2\pi {\epsilon }_0}\frac{\lambda R}{R^2}{\int }_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0 cos \theta d\theta \\ \Rightarrow E=\frac{1}{2\pi {\epsilon }_0}\frac{R}{R^2}\left(\frac{Q}{{\displaystyle \raisebox{1ex}{$\pi R$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right){\left[\sin \theta \right]}_{\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}^0 \\ \Rightarrow E=\frac{1}{{\pi }^2{\epsilon }_0}\frac{Q}{R^2}\left[0-1\right] \\ \Rightarrow E=\frac{-1}{{\pi }^2{\epsilon }_0}\frac{Q}{R^2} \\ \Rightarrow \left|E\right|=\frac{1}{{\pi }^2{\epsilon }_0}\frac{Q}{R^2} \end{gathered}$$

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