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A thin glass rod is bend 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 at P, the centre of 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

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$dE = \frac{1}{4 {πε}_{0}} \frac{Qrdθ}{r^{2}} / (\frac{πr}{2}) = \frac{Qdθ}{2 π^{2} {εr}^{2}}$

${dE}_{x} = \frac{Q}{2 {πε}_{0} r^{2}} \cos θdθ$

${dE}_{y} = \frac{Q}{2 {πε}_{0} r^{2}} \sin θdθ$

X component get cancelled net electric field along y direction

$\int d E_{y} = \int_{0}^{\frac{π}{2}} \frac{Q}{2 {πε}_{0} r^{2}} \sin θdθ$

$E_{y} = \frac{Q}{π^{2} ε_{0} r^{2}}$

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