AP® Physics C: Electricity and Magnetism · Unit 8 · Topic 8.4
The Field of a Charged Rod
A charged rod's field comes from adding up tiny point-charge contributions along its length with an integral; the answer depends on geometry, and on the bisector it simplifies because sideways components cancel.
A rod is not a point charge, so you can't just plug into kQ/r2. Instead you chop the rod into tiny pieces of charge dq, treat each as a point charge, write its tiny field dE = k dq / r2, and add up all the pieces with an integral. The setup depends on where your point is. On the rod's own axis, beyond one end, every dq points its field in the same direction, so the integral is scalar and telescopes neatly to E = kQ / (x(x+L)), where x is the distance from the nearer end and L is the rod's length. On the perpendicular bisector, each dq's field has a component along the rod and a component perpendicular to it; by symmetry the along-rod components from charge on opposite sides cancel, leaving only the perpendicular piece to integrate.
Close to a long rod, the field falls off like 1/r instead of 1/r2. A point charge's field thins out in all directions as you retreat, but near a long rod, retreating also brings more of the rod's distant charge into effective range, partially compensating for the usual 1/r2 drop. The net result is a milder 1/r falloff, a signature you can spot immediately on any long charged conductor.
This topic is the proving ground for Gauss's law. Once you see how painful the direct integral gets for an infinite rod, Gauss's law using a cylindrical surface will feel like a shortcut you can't live without, built on the exact symmetry argument you used here to kill the sideways field component.
Key terms
- Linear charge density
- Charge per unit length along the rod, often written lambda = Q/L.
- dq
- An infinitesimal slice of the rod's charge, treated as a point charge inside the integral.
- Perpendicular bisector
- The line through the rod's midpoint, perpendicular to the rod, where sideways field components cancel by symmetry.
- Telescoping integral
- An integral that simplifies dramatically because its antiderivative evaluated at the limits collapses to a simple expression.
Watch it explained
- How it worksWalk me through the integral for the field on the perpendicular bisector of a charged rod.
- Worked exampleFind the field 5 cm past the end of a 20 cm rod carrying 12 nC, along the rod's own axis.
- Common mistakeWhy doesn't the sideways component survive on the bisector of a symmetric charged rod?
- Going furtherWhat does the rod's field become when its length runs off toward infinity?
Check yourself
End of topic.