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joe_cool2
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Homework Statement
FIGURE P27.43 shows a thin rod of length L with total charge Q. Find an expression for the electric field at distance x from the rod. Give your answer in component form.
Homework Equations
Equation for an electric field:
E= \frac{KQ}{r^2}
The Attempt at a Solution
Exi = \frac {1}{4\pi\epsilon_0} \frac{Q}{r^2} cos\theta
Ex = \frac{Q}{4\pi\epsilon_0} \sum_{i=1}^n \frac{1}{x^2+y^2} \frac{x}{\sqrt{x^2+y^2}}
Let's take the limit as n approaches infinity, thereby getting us an integral...
\int \frac{x}{(x^2 + y^2)^\frac{3}{2}} dy
x \int \frac{dy}{(x^2 + y^2)^\frac{3}{2}}
\frac{x^2}{x^3} \int \frac{dy}{(x^2 + y^2)^\frac{3}{2}}
\frac{sin\theta}{x} + C
\frac{y}{x\sqrt{x^2+y^2}} + C
The actual x-component of the electric field should be this indefinite integral evaluated from 0 to L and multiplied by Q over 4 times pi times epsilon nought...
Ex = \frac {Q}{4\pi\epsilon_0} \frac {L}{x\sqrt{x^2+L^2}}
The y-component may be evaluated in a similar manner.
Ey = \frac{Q}{4\pi\epsilon_0} \sum_{i=1}^n \frac{1}{x^2+y^2} \frac{y}{\sqrt{x^2+y^2}}
Ey = \frac{Q}{4\pi\epsilon_0} \sum_{i=1}^n \frac{1}{x^2+y^2} \frac{y}{\sqrt{x^2+y^2}}
\int \frac{y}{(x^2 + y^2)^\frac{3}{2}} dy
\frac{x^2}{x^3} \int \frac{tan\theta sec^2\theta d\theta}{sec^3\theta}
\frac{-cos\theta}{x} + C
\frac{-1}{\sqrt{x^2 + y^2}}
y-component:
Ey = \frac {Q}{4\pi\epsilon_0} \left(\frac {1}{\sqrt{x^2+L^2}} - \frac {1}{x} \right)
According to the odd answers in the back of my textbook, this is not correct!
So I did some even problems before this that were almost as complex, but not quite. I'd feel silly posting three separate topics, so perhaps I'll post them in this thread after this one is answered? According to the table of integrals in the back of my book, it looks like I'm doing it right, so maybe I set it up wrong?
