The electric field at the center of a uniformly charged half ring is given by:
E=R2kλ
where λ=πRQ is the linear charge density.
Substituting λ, we get:
E=πR22kQ=4πϵ0πR22Q=2π2ϵ0R2Q
Rearranging for Q:
Q=2π2ϵ0R2E
Given values:
E=100 V/m
R=35 cm =0.35 m
ϵ0=8.85×10−12 C2/Nm2
π=3.14
Substituting the values:
Q=2×(3.14)2×8.85×10−12×(0.35)2×100
Q=2×9.8596×8.85×10−12×0.1225×100
Q≈2137.8×10−12 C
Q≈2.14×10−9 C =2.14 nC
Answer: 2.14