Calculate Total Charge on Uniformly Charged Half Ring

To find the total charge $Q$ on the thin half ring, we first write the expression for the electric field at the center of a uniformly charged circular arc.

For a half ring (a circular arc of angle $\theta = \pi$) of radius $R$ and total charge $Q$, the linear charge density $\lambda$ is given by: $\lambda = \frac{Q}{\pi R}$

The magnitude of the electric field $E$ at the center of a half ring is: $E = \frac{2 k \lambda}{R}$

Substituting $k = \frac{1}{4\pi\epsilon_0}$ and $\lambda = \frac{Q}{\pi R}$ into the equation: $E = \frac{2 \left(\frac{1}{4\pi\epsilon_0}\right) \left(\frac{Q}{\pi R}\right)}{R} = \frac{Q}{2\pi^2\epsilon_0 R^2}$

Rearranging the formula to solve for the total charge $Q$: $Q = 2\pi^2\epsilon_0 R^2 E$

Given data:

• Radius, $R = 35\text{ cm} = 0.35\text{ m}$
• Magnitude of Electric field, $E = 100\text{ V/m}$
• Permittivity of free space, $\epsilon_0 = 8.85 \times 10^{-12}\text{ C}^2/\text{Nm}^2$
• $\pi = 3.14$

Substitute these values into the expression for $Q$: $Q = 2 \times (3.14)^2 \times (8.85 \times 10^{-12}) \times (0.35)^2 \times 100$

Calculating step-by-step: $\pi^2 \approx 3.14^2 = 9.8596$ $R^2 = (0.35)^2 = 0.1225\text{ m}^2$

Now, substituting these back into the calculation: $Q = 2 \times 9.8596 \times 8.85 \times 10^{-12} \times 0.1225 \times 100$ $Q = 24.5 \times 9.8596 \times 8.85 \times 10^{-12}$ $Q = 2137.81 \times 10^{-12}\text{ C}$ $Q = 2.13781 \times 10^{-9}\text{ C} \approx 2.14\text{ nC}$

Thus, the value of $Q$ is $2.14\text{ nC}$.