Electric field The magnitude of the electric field at a point x meters from the midpoint of a 0.1-m line of charge is given by E(x)=4.35xx2+0.01 (in units of newtons per coulomb, N/C). Evaluate limx→10E(x).
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\[ E(x) = \frac{4.35x}{x^2 + 0.01} \] Show more…
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The magnitude of the electric field at a point $x$ meters from the midpoint of a 0.1 -m line of charge is given by $E(x)=\frac{4.35}{x \sqrt{x^{2}+0.01}}(\text { in units of newtons per coulomb, } \mathrm{N} / \mathrm{C}).$ Evaluate $\lim _{x \rightarrow 10} E(x).$
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Electric field The magnitude of the electric field at a point $x$ meters from the midpoint of a 0.1 -m line of charge is given by $\left.E(x)=\frac{4.35}{x \sqrt{x^{2}+0.01}} \text { (in units of newtons per coulomb, } \mathrm{N} / \mathrm{C}\right)$ Evaluate $\lim _{x \rightarrow 10} E(x)$
A charged line like that shown in Fig. 21.25 extends from $y=2.50 \mathrm{cm}$ to $y=-2.50 \mathrm{cm} .$ The total charge distributed uniformly along the line is $-9.00 \mathrm{nC}$ . (a) Find the electric field (magnitude and direction) on the $x$ -axis at $x=10.0 \mathrm{cm} .$ (b) Is the magnitude of the electric field you calculated in part (a) larger or smaller than the electric field 10.0 $\mathrm{cm}$ from a point charge that has the same total charge as this finite line of charge? In terms of the approximation used to derive $E=Q / 4 \pi \epsilon_{0} x^{2}$ for a point charge from Eq. $(21.9),$ explain why this is so. (c) At what distance $x$ does the result for the finite line of charge differ by 1.0$\%$ from that for the point charge?
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