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The formula for the electric field at a point due to a charge $Q$ (just considering the magnitude) at some distance $x$ away from the point is $E=\dfrac{k_eQ}{x^2}$ where $k_e$ is a constant equal to approximately $8.99 \times 10^{9}$.

If we now consider a uniformly charged rod with charge density $\lambda$ and length $L$, then the charge of the entire rod is $\lambda L$.

Assuming a point $P$ that is $a$ units from the end of the rod along the $x$-axis, how would I express the derivative of the electric field with respect to $x$ when the charge $Q$ is a function of $x$, but the electric field $E$ varies with both $Q$ and the inverse square of the distance between $x$ and the point $P$? Isn't $E$ a function of two variables in this case?

Ng Chung Tak
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julieb
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1 Answers1

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Let's write $Q$ as a function of distance $R$ from the particle: $$Q=\int_x^R \lambda dR$$ This tells us that $$\frac{dQ}{dR}=\lambda\to dQ=\lambda dR\tag{1}$$ Now, $$E=k_e\frac{Q}{R^2}\to E=\int_0^EdE=\int_0^Qk_e\frac{dQ}{R^2}\tag{2}$$ Substitution of $(1)$ into $(2)$ yields $$E=k_e\int_x^{x+L}\frac{\lambda}{R^2}dR$$ Can you go from here?

HDE 226868
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  • How did you know to take this approach vs. starting with the equation for E and differentiating directly? Is this sort of a "trick" that comes with experience? How did you know in equation 2 to use dQ in the numerator but not the derivative of R squared in the denominator? – julieb Sep 01 '16 at 21:10
  • @julieb Trial and error, in the past. So yes, I guess you could call it "a trick that comes with experience". – HDE 226868 Sep 02 '16 at 15:11
  • Hmmm, I'm still hung up. Let's assume that the rod's length is along the x-axis, that it's length is L, and the distance to the end of the rod to the point P is "a". The amount of charge Q in a 1-D rod is a function of dx, because it's equal to charge density times the amount of rod you're evaluating at. So it's like Q = Q(dx). Then you have E, whose strength at each point x along the rod varies by the inverse of ${L+a - x}^{2}$. So if I want to find the derivative of E with respect to x, would I use the chain rule or something? – julieb Sep 02 '16 at 19:51
  • I meant $(L + a -x)^2$ – julieb Sep 02 '16 at 20:00
  • @julieb $Q$ isn't a function of $dx$; it's a function of the limits of the integral $Q=\int_a^{a+x}\lambda dx$. If $a=L$, then the charge is the total charge on the rod, $\lambda L$. – HDE 226868 Sep 03 '16 at 01:20