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Chapter 26: Q.15 (page 738)

The electric potential along the x-axis is $V = 100 e^{- 2 x} V$ , where $x$ is in meters. What is $E_{x}$ at (a)role="math" localid="1649581712352" $x = 1.0 m$ and (b) role="math" localid="1649581719374" $x = 2.0 m$ ?

Short Answer

Expert verified

$(a) E_{x} = 27 V / m (b) E_{x} = 3.7 V / m$

Step by step solution

01

Given information and theory used

Given : The electric potential along the x-axis is : $V = 100 e^{- 2 x} V$

(a) $x = 1 . 0 m$ and (b) $x = 2.0 m$

Theory used :

The electric field is expressed as a function of potential as :

$E (x) = - \frac{d V (x)}{d x}$

02

Calculating Ex

The negative derivative of the potential $V = 100 e^{- 2 x} V$ is equivalent to the component of the electric field. As a result,

role="math" localid="1649581602113" $E (x) = - \frac{d V (x)}{d x} = - 100 (- 2 e^{- 2 x}) V / m = 200 e^{- 2 x} V / m$

(a) We have $E_{x} = 27 V / m at a = 1.0 m$ .

(a) We have $E_{x} = 3.7 V / m at x = 2.0 m$ .

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Most popular questions from this chapter

What are the charge on and the potential difference across each capacitor in FIGURE $P 26.57$ ?

Two 3.0-cm-diameter aluminum electrodes are spaced 0.50 mm apart. The electrodes are connected to a 100 V battery.

a. What is the capacitance?

b. What is the magnitude of the charge on each electrode?

What is the potential difference $鈭� V_{34}$ in Figure EX26.16?

a. Use the methods of Chapter 25 to find the potential at distance $x$ on the axis of the charged rod shown in FIGURE P26.43.

b. Use the result of part a to find the electric field at distance $x$ on the axis of a rod

Estimate the electric fields ${\overset{鈫�}{E}}_{1}$ and ${\overset{鈫�}{E}}_{2}$ at points 1 and 2 in Figure Q26.4. Don鈥檛 forget that $\overset{鈫�}{E}$ is a vector.

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