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. (\(\epsilon_o = 8.85 \times 10^{-12}\) C\(^2\)/Nm\(^2\) and \(\pi = 3.14\))

The electric field at the center of a uniformly charged half ring is given by:

\(E = \dfrac{2k\lambda}{R}\)

where \(\lambda = \dfrac{Q}{\pi R}\) is the linear charge density.

Substituting \(\lambda\), we get:

\(E = \dfrac{2kQ}{\pi R^2} = \dfrac{2Q}{4\pi\epsilon_0 \pi R^2} = \dfrac{Q}{2\pi^2\epsilon_0 R^2}\)

Rearranging for \(Q\):

\(Q = 2\pi^2\epsilon_0 R^2 E\)

Given values:
\(E = 100\) V/m
\(R = 35\) cm \(= 0.35\) m
\(\epsilon_0 = 8.85 \times 10^{-12}\) C\(^2\)/Nm\(^2\)
\(\pi = 3.14\)

Substituting the values:

\(Q = 2 \times (3.14)^2 \times 8.85 \times 10^{-12} \times (0.35)^2 \times 100\)

\(Q = 2 \times 9.8596 \times 8.85 \times 10^{-12} \times 0.1225 \times 100\)

\(Q \approx 2137.8 \times 10^{-12}\) C

\(Q \approx 2.14 \times 10^{-9}\) C \(= 2.14\) nC

Answer: \(2.14\)