5. Consider an infinitely thin charged rod of length L with uniform linear charg

Question

5. Consider an infinitely thin charged rod of length L with uniform linear charge density lambda lying along the x-axis. The ends are at x = +-L/2 and the center is at the origin. a. Calculate the electric field (magnitude and direction) for x > L/2. b. Calculate the force on a test particle with charge q0 L/2. c. Show that the force on the test particle will simplify to Coulomb's Law when it is sufficiently far away from the rod, i.e. when x >> L. Use Q = Lambda L for the total charge on the rod.

Explanation / Answer

Solution:

Consider an uniformly charged wire of infinite length having a constant linear charge density ? (charge per unit length). Let P be a point at a distance x from the rod (where x = L/2) and E be the electric field at the point P. A cylinder of length L, radius L/2, closed at each end by plane caps normal to the axis is chosen as Gaussian surface. Consider a very small area ds on the Gaussian surface.

The magnitude of the electric field will be the same at all points on the curved surface of the cylinder and directed radially outward. E and ds are along the same direction. The electric flux (?) through curved surface
? = closed ? E ds cos ?
[ ? ? = 0;cos ? = 1 ]
? = E ds
?- Electric flux
? = E ds = E (2?xl)
( ? The surface area of the curved part is 2? xl)
Since E and ds are right angles to each other, the electric flux
through the plane caps = 0
?
Total flux through the Gaussian surface, ? = E. (2?xl)
The net charge enclosed by Gaussian surface is, q = ?l
?
By Gauss

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