GREPhysics.NET: GR8677 #98

GR8677 #98

Problem

GREPhysics.NET Official Solution

Electromagnetism $\Rightarrow$ }Potential

Recall that $V=\int \frac{dq}{x}=k\int_l^{2l} \lambda \frac{dx}{x}=k\lambda\ln 2$ , where $\lambda=Q/l$ is your usual linear charge density.

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Comments

rohit 2008-11-07 07:52:44	firstly , the limits are from 0 to l ( over the charge distribution ) second, the the integrand is dq/(x+l) Reply to this comment
a19grey2 2008-11-04 16:41:12	To clarify, the correct answer is D. Reply to this comment
Andresito 2006-03-21 00:48:48	V = k * integral (dq / x) , in the first expression. Reply to this comment

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V = k * integral (dq / x) , in the first expression.

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

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Bare Basic LaTeX Rosetta Stone