Positive electric charge Q is distributed uniformly along a thin rod of length 2a. The rod lies along the x-axis between x=−a and x=+a (Figure 1). Calculate how much work you must do to bring a positive point charge q from infinity to the point x=+L on the x-axis, where L>a. Positive electric charge is distributed uniformly along athin rod of length 2a. The rod lies along the x-axisbetween x = -a and x = +a (Figure 1). Calculatehow much work you must do to bring a positive pointcharge q from infinity to the point x = +L on the x-axis,where L > a.Figurex=-aRod with charge Qx=0x = +a< 1 of 1Point charge qbrought from infinityx = +LxIn this problem you must first calculate the potential V at x = +L due to the charged rod. You can thenfind the change in potential energy involved in bringing the point charge q from infinity (where V = 0) tox = +L.Part ATo find V, divide the rod into infinitesimal segments of length dx₁. Consider one such segment locatedat x = x₁, where -a ≤ x₁ ≤ a. What is the potential dV at x = +L due to this segment?Express your answer in terms of some or all of the variables Q, a, L, dx₁, x₁, electric constant€0, and T.View Available Hint(s)Hint 1. The charge of a segment of infinitesimal length.VE ΑΣΦВαVRA ΣYPΦ86Y€TΩnΦħᎾXEK=λY @XReview | Constants?μwww

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College Physics

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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Positive electric charge Q is distributed uniformly along a thin rod of length 2a. The rod lies along the x-axis between x=−a and x=+a (Figure 1). Calculate how much work you must do to bring a positive point charge q from infinity to the point x=+L on the x-axis, where L>a.

Positive electric charge is distributed uniformly along a
thin rod of length 2a. The rod lies along the x-axis
between x = -a and x = +a (Figure 1). Calculate
how much work you must do to bring a positive point
charge q from infinity to the point x = +L on the x-axis,
where L > a.
Figure
x=-a
Rod with charge Q
x=0
x = +a
< 1 of 1
Point charge q
brought from infinity
x = +L
x
In this problem you must first calculate the potential V at x = +L due to the charged rod. You can then
find the change in potential energy involved in bringing the point charge q from infinity (where V = 0) to
x = +L.
Part A
To find V, divide the rod into infinitesimal segments of length dx₁. Consider one such segment located
at x = x₁, where -a ≤ x₁ ≤ a. What is the potential dV at x = +L due to this segment?
Express your answer in terms of some or all of the variables Q, a, L, dx₁, x₁, electric constant
€0, and T.
View Available Hint(s)
Hint 1. The charge of a segment of infinitesimal length.
VE ΑΣΦ
В
α
V
R
A Σ
Y
P
Φ
8
6
Y
€
T
Ω
n
Φ
ħ
Ꮎ
X
E
K
=
λ
Y @
X
Review | Constants
?
μ
www

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What does your result for the potential energy U(x=+L) become in the limit a→0?

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