# Electric Flux Through a Closed Cylinder (Step-by-Step Solution) Let’s solve the problem step by step. --- ## **1. **Given Data** - The cylinder contains a line charge along its axis. - Linear charge density: \(\lambda = \lambda_ \sin\left(\frac{2\pi z}{h}\right)\) - Cylinder (and line charge) extends from \(z = \) to \(z = h/2\). --- ## **2. **Find the Total Enclosed Charge \(Q_{enc}\)** The total charge enclosed by the cylinder is the integral of the linear charge density over the length: \[ Q_{enc} = \int_{z=}^{z=h/2} \lambda(z)\, dz \] Substitute the given \(\lambda(z)\): \[ Q_{enc} = \int_^{h/2} \lambda_ \sin\left(\frac{2\pi z}{h}\right) dz \] --- ## **3. **Evaluate the Integral** Let’s solve the integral: \[ Q_{enc} = \lambda_ \int_^{h/2} \sin\left(\frac{2\pi z}{h}\right) dz \] Let \(u = \frac{2\pi z}{h}\), so \(dz = \frac{h}{2\pi} du\). When \(z = \rightarrow u = \) When \(z = h/2 \rightarrow u = \pi\) So, \[ Q_{enc} = \lambda_ \int_{u=}^{u=\pi} \sin(u) \cdot \frac{h}{2\pi} du = \lambda_ \frac{h}{2\pi} \int_^{\pi} \sin(u) du \] \[ \int_^{\pi} \sin(u) du = [-\cos(u)]_^{\pi} = -\cos(\pi) + \cos() = -(-1) + 1 = 2 \] So, \[ Q_{enc} = \lambda_ \frac{h}{2\pi} \times 2 = \lambda_ \frac{h}{\pi} \] --- ## **4. **Apply Gauss's Law for Electric Flux** By Gauss’s Law: \[ \Phi_E = \frac{Q_{enc}}{\epsilon_} \] Substitute the value of \(Q_{enc}\): \[ \Phi_E = \frac{\lambda_ h}{\pi \epsilon_} \] --- ## **Final Answer** \[ \boxed{ \Phi_E = \frac{\lambda_ h}{\pi \epsilon_} } \] This is the total electric flux through the closed cylinder.