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 Fig. 1.133. Calculate electric field $E$ at $P$, the center of semicircle.
Fig. $1.133$
a. $\frac{Q}{\pi^{2} \varepsilon_{0} r^{2}}$
b. $\frac{2 Q}{\pi^{2} \varepsilon_{0} r^{2}}$
c. $\frac{4 Q}{\pi^{2} \varepsilon_{0} r^{2}}$
d. $\frac{Q}{4 \pi^{2} \varepsilon_{0} r^{2}}$