A thin half ring of radius 35 cm is uniformly charged with a total...

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: $2.14$