Question

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}}\\) (in units of newtons per coulomb, n/c). evaluate

\\\lim_{x \to 18} e(x)\\

\\\lim_{x \to 18} e(x) \approx \text{n/c}\\
(round to four decimal places as needed.)

Explanation:

Substitute the limit value into the function

$$ E(18) = \frac{4.35}{18\sqrt{18^2 + 0.01}} $$

Calculate the denominator and evaluate the expression

$$
LATEXBLOCK0
$$

Round to four decimal places

$$ \lim_{x \to 18} E(x) \approx 0.0134 $$

Answer:

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 \(E(x) = \frac{4.35}{x\sqrt{x^2 + 0.01}}\) (in units of newtons per coulomb, N/C). Evaluate \(\lim_{x \to 18} E(x)\).

\(\lim_{x \to 18} E(x) \approx\) <blank>0.0134</blank> N/C
(Round to four decimal places as needed.)