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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

A nerve cell in its resting state has a membrane potential of $- 70 mV$ , meaning that the potential inside the cell is $70 mV$ less than the potential outside due to a layer of negative charge on the inner surface of the cell wall and a layer of positive charge on the outer surface. This effectively makes the cell wall a charged capacitor. When the nerve cell fires, sodium ions, ${Na}^{+}$ , flood through the cell wall to briefly switch the membrane potential to $+ 40 mV$ . Model the central body of a nerve cell-the soma-as a $50 渭 m$ diameter sphere with a $7.0 - nm$ -thick cell wall whose dielectric constant is 9.0. Because a cell's diameter is much larger than the wall thickness, it is reasonable to ignore the curvature of the cell and think of it as a parallel-plate capacitor. How many sodium ions enter the cell as it fires?

A switch that connects a battery to a 10 mF capacitor is closed. Several seconds later you find that the capacitor plates are charged to $卤$ 30 mC. What is the emf of the battery?

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.

Consider a uniformly charged sphere of radius R and total cAlC charge Q. The electric field $E_{out}$ outside the sphere $(r 鈮� R)$ is simply that of a point charge Q. In Chapter 24, we used Gauss's law to find that the electric field $E_{in}$ inside the sphere $(r 鈮� R)$ is radially outward with field strength

$E_{in} = \frac{1}{4 蟺系_{0}} \frac{Q}{R^{3}} r$

a. The electric potential $V_{out}$ outside the sphere is that of a point charge Q. Find an expression for the electric potential $V_{in}$ at position r inside the sphere. As a reference, let $V_{in} = V_{out}$ at the surface of the sphere.

b. What is the ratio $V_{center} / V_{surface} ?$

c. Graph V versus r for 0 $鈮�$ r $鈮�$ 3 R.

A $2.0 c m$ -diameter parallel-plate capacitor with a spacing of $0.50 m m$ is charged to $200 V$ . What are

(a) the total energy stored in the electric field and

(b) the energy density?

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