We are going to calculate the on-axis electric field due to a rod with uniform linear charge density. We will assume that a rod of length L and linear charge density A lies along the z-axis. We will calculate the electric field at point P, which is along the axis of the rod and a distance d from its left end. Po X * d L dQ Our equation sheet tells us that d = K -. In this problem, every piece of differential electric 7.2 field de points in the same direction (to the left when A is positive). Because of this we can ignore direction in this problem and just say that E dE, and treat everything as positive inside the - [dE integral. Therefore E dQ 72 Step 1: In the box below, click Vars and then choose the combination of these variables that gives us the differential charge dQ. dQ

Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter19: Electric Forces And Electric Fields
Section: Chapter Questions
Problem 73P: Two infinite, nonconducting sheets of charge are parallel to each other as shown in Figure P19.73....
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We are going to calculate the on-axis electric field due to a rod with uniform linear charge density.
We will assume that a rod of length L and linear charge density A lies along the x-axis. We will
calculate the electric field at point P, which is along the axis of the rod and a distance d from its left
end.
P.
X
* d
dQ
Our equation sheet tells us that d = K
-. In this problem, every piece of differential electric
7.2
field de points in the same direction (to the left when A is positive). Because of this we can ignore
dE, and treat everything as positive inside the
- fdE
direction in this problem and just say that E
integral. Therefore E
L
dQ
72
Step 1: In the box below, click Vars and then choose the combination of these variables that gives us
the differential charge dQ.
dQ
Transcribed Image Text:We are going to calculate the on-axis electric field due to a rod with uniform linear charge density. We will assume that a rod of length L and linear charge density A lies along the x-axis. We will calculate the electric field at point P, which is along the axis of the rod and a distance d from its left end. P. X * d dQ Our equation sheet tells us that d = K -. In this problem, every piece of differential electric 7.2 field de points in the same direction (to the left when A is positive). Because of this we can ignore dE, and treat everything as positive inside the - fdE direction in this problem and just say that E integral. Therefore E L dQ 72 Step 1: In the box below, click Vars and then choose the combination of these variables that gives us the differential charge dQ. dQ
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