A thin glass rod is bent into a semicircle of radius r. A charge +q is uniformly distributed
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Answer:
Take PO as the x-axis and PA as the y-axis. Consider two elements EF and E'F' of width dθ at angular distance θ above and below PO, respectively.
The magnitude of the field at P due to either element is
dE=
4πε
0
1
r
2
rdθ×Q/(πr/2)
=
2π
2
ε
0
r
2
Q
dθ
Resolving the fields, we find that the components along PO sum up to zero, and hence the resultant field is along PA.
Therefore, field at P due to pair of elements is 2 E sin θ
E=∫
0
π/2
2Esinθ
=2∫
0
π/2
2π
2
ε
0
r
2
Q
sinθdθ=
π
2
ε
0
r
2
Q
Explanation:
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