A thin half ring of radius 35 cm is uniformly charged with a total charge of Q coulomb. If the magnitude of the electric field at centre of the half ring is 100 V/m, then the value of Q is _______ nC. ( _o = 8.85 10⁻¹² C ^2 /Nm ^2 and = 3.14 )

Step-by-step solution

The electric field at the center of a uniformly charged half ring is given by: E = 2k R where = Q R is the linear charge density. Substituting , we get: E = 2kQ R^2 = 2Q 4 ₀ R^2 = Q 2 ^2 ₀ R^2 Rearranging for Q : Q = 2 ^2 ₀ R^2 E Given values: E = 100 V/m R = 35 cm = 0.35 m ₀ = 8.85 10⁻¹² C ^2 /Nm ^2 = 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