Electric Field Calculation Using Gauss's Law

To determine the electric field outside a given cylindrical configuration, we can effectively use Gauss's Law. This problem involves a long, straight wire surrounded by a hollow metal cylinder, both having specific charge per unit length values.

Understanding the Setup

• A long, straight wire is placed along the axis.
• The wire has a charge per unit length, denoted as \(\lambda\).
• A hollow metal cylinder surrounds the wire, with its axis coinciding with that of the wire.
• The cylinder has a net charge per unit length of \(2\lambda\).
• We need to find the electric field at a distance \(r\) from the axis, where \(r\) is outside the cylinder.

Applying Gauss's Law for Cylindrical Symmetry

Due to the cylindrical symmetry of the charge distribution (a long wire and a coaxial cylinder), the electric field will be radial and its magnitude will depend only on the distance \(r\) from the axis. To apply Gauss's Law, we choose a cylindrical Gaussian surface.

1. Gaussian Surface Selection:

   Consider a cylindrical Gaussian surface of radius \(r\) (where \(r\) is outside the hollow metal cylinder) and arbitrary length \(L\). This Gaussian surface is coaxial with the wire and the hollow cylinder.

2. Gauss's Law Statement:

   Gauss's Law states that the total electric flux \(\Phi_E\) through any closed surface is equal to the total enclosed charge \(Q_{enc}\) divided by the permittivity of free space \(\varepsilon_0\).

   Mathematically, it is expressed as:

   \[ \oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\varepsilon_0} \]

3. Calculating Electric Flux:

   For our chosen cylindrical Gaussian surface:

   • The electric field \(\vec{E}\) is radial and perpendicular to the curved surface of the cylinder. Thus, \(\vec{E}\) is parallel to the area vector \(d\vec{A}\) on the curved surface.
   • The magnitude of the electric field \(E\) is constant over the curved surface at a fixed radius \(r\).
   • The area of the curved surface of the Gaussian cylinder is \(2\pi r L\).
   • The electric field lines are parallel to the end caps of the Gaussian cylinder, meaning the flux through the end caps is zero.

   Therefore, the total electric flux \(\Phi_E\) simplifies to:

   \[ \Phi_E = E \cdot (2\pi r L) \]

4. Determining Enclosed Charge \(Q_{enc}\):

   The Gaussian surface is positioned *outside* the hollow metal cylinder. This means it encloses all the charge from both the inner wire and the hollow metal cylinder.

   • Charge enclosed from the long straight wire (charge per unit length \(\lambda\)): \(Q_{wire} = \lambda L\)
   • Charge enclosed from the hollow metal cylinder (net charge per unit length \(2\lambda\)): \(Q_{cylinder} = 2\lambda L\)

   The total enclosed charge \(Q_{enc}\) is the sum of these charges:

   \[ Q_{enc} = Q_{wire} + Q_{cylinder} = \lambda L + 2\lambda L = 3\lambda L \]

5. Solving for the Electric Field \(E\):

   Now, substitute the expressions for flux and enclosed charge back into Gauss's Law:

   \[ E \cdot (2\pi r L) = \frac{3\lambda L}{\varepsilon_0} \]

   To find \(E\), divide both sides by \(2\pi r L\):

   \[ E = \frac{3\lambda L}{2\pi r L \varepsilon_0} \]

   The length \(L\) cancels out:

   \[ E = \frac{3\lambda}{2\pi \varepsilon_0 r} \]

Direction of the Electric Field

Since the total enclosed charge \(3\lambda L\) is positive (assuming \(\lambda\) is a positive charge density), the electric field lines will point away from the source charges. For a cylindrically symmetric positive charge distribution, this means the electric field is directed radially outward.

Thus, the electric field outside the cylinder, at a distance \(r\) from the axis, is \(\frac{3\lambda}{2\pi \varepsilon_0 r}\) radially outward.