An infinite line of charge with uniform charge density λ is enclosed by a cylinder of length L and radius R. How much charge is enclosed by the cylinder?
The correct answer and explanation is:
To find the charge enclosed by the cylinder, we can use Gauss’s Law, which is a fundamental principle in electromagnetism. Gauss’s Law states that the electric flux through a closed surface is directly proportional to the charge enclosed by that surface. The equation is: ΦE=Qencϵ0\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}
Where:
- ΦE\Phi_E is the electric flux through the surface.
- QencQ_{\text{enc}} is the charge enclosed by the surface.
- ϵ0\epsilon_0 is the permittivity of free space.
Steps for solving:
- Choose a Gaussian surface: In this case, the Gaussian surface is a cylindrical surface that surrounds the infinite line of charge. The cylinder is concentric with the line of charge. This is chosen because the symmetry of the problem (an infinite line charge) makes the electric field at a given radial distance rr from the axis of the line of charge only depend on rr and point radially outward.
- Apply Gauss’s Law: The electric flux ΦE\Phi_E through the cylindrical surface can be expressed as:
ΦE=E⋅A\Phi_E = E \cdot A
Where:
- EE is the electric field at a distance rr from the line of charge.
- AA is the lateral surface area of the Gaussian cylinder, which is 2πrL2\pi rL, where LL is the length of the cylinder and rr is the radial distance from the line of charge.
- Relate electric field to charge: According to Gauss’s Law:
ΦE=Qencϵ0\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}
For the infinite line of charge, the electric field at distance rr from the line is given by: E=2keλrE = \frac{2k_e \lambda}{r}
Where:
- kek_e is Coulomb’s constant (8.99×109 N m2/C28.99 \times 10^9 \, \text{N m}^2/\text{C}^2).
- λ\lambda is the linear charge density (charge per unit length) of the line.
- Calculate the charge enclosed: Now, substitute the expression for the electric field into Gauss’s Law. The total electric flux is:
ΦE=E⋅A=(2keλr)⋅(2πrL)=4πkeλLr\Phi_E = E \cdot A = \left(\frac{2k_e \lambda}{r}\right) \cdot (2\pi rL) = \frac{4\pi k_e \lambda L}{r}
From Gauss’s law, the flux is also equal to: ΦE=Qencϵ0\Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0}
Equating the two expressions for flux: 4πkeλLr=Qencϵ0\frac{4\pi k_e \lambda L}{r} = \frac{Q_{\text{enc}}}{\epsilon_0}
Solving for the enclosed charge QencQ_{\text{enc}}, we get: Qenc=4πkeλLϵ0Q_{\text{enc}} = 4\pi k_e \lambda L \epsilon_0
Since ke=14πϵ0k_e = \frac{1}{4\pi \epsilon_0}, the expression simplifies to: Qenc=λLQ_{\text{enc}} = \lambda L
Final Answer:
The charge enclosed by the cylinder is Qenc=λLQ_{\text{enc}} = \lambda L, where λ\lambda is the linear charge density and LL is the length of the cylinder.
This result shows that the enclosed charge depends directly on the length of the cylinder and the charge density of the line. The radius RR of the cylinder does not affect the enclosed charge since the charge is distributed along the infinite line.