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 figure. The electric field E at P, the center of the semicircle, is

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$\frac{Q}{π^{2} ε_{0} r^{2}}$

$\frac{2 Q}{π^{2} ε_{0} r^{2}}$

$\frac{4 Q}{π^{2} ε_{0} r^{2}}$

$\frac{Q}{4 π^{2} ε_{0} r^{2}}$

answer is A.

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Detailed Solution

Take PO as the x-axis and PA as the y-axis. Consider two elements EF and E'F' of width $d θ$ at angular distance $θ$ above and below PO, respectively.

The magnitude of the field at P due to either element is

$d E = \frac{1}{4 π ε_{0}} \frac{r d θ \times Q / (π r / 2)}{r^{2}} = \frac{Q}{2 π^{2} ε_{0} r^{2}} d θ$

Resolving the fields, we find that the components along PO sum up to zero, and hence the resultant field is along PB. Therefore, field at P due to pair of elements is 2dEsin $θ$

$E = \int_{0}^{π / 2} 2 d E \sin θ = 2 \int_{0}^{π / 2} \frac{Q}{2 π^{2} ε_{0} r^{2}} \sin θ d θ = \frac{Q}{π^{2} ε_{0} r^{2}}$