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Chapter 24: Q72P (page 715)

The magnitude $E$ of an electric field depends on the radial distance r according to $E = A / r^{4}$ , where is a constant with the unit volt鈥揷ubic meter. As a multiple of $A$ , what is the magnitude of the electric potential difference between $r = 2.00 m$ and $r = 3.00 m$ ?

Short Answer

Expert verified

The magnitude of the electric potential difference is $V_{(r)} = + 2.93 脳 10^{鈭� 2} A$ .

Step by step solution

01

Given data:

The intensity of the electric field $E = \frac{A}{r^{4}}$ ,

The value of distance are $r_{1} = 2.0 m$ and $r_{2} = 3.0 m$

02

Understanding the concept

Use the relation between the electric field $E$ and potential $V$ as given below.

role="math" localid="1662730170007" $E = \frac{鈭� dV}{dr}$

Here, $r$ is the distance.

03

Calculate the magnitude of the electric potential difference between r= 2.00 m and r= 3.00 m:

Since, the electric field is,

$\begin{matrix} E = \frac{鈭� dV}{dr} \\ {鈭�}_{鈭�}^{r} dV = 鈭� {鈭�}_{畏}^{n} Edr \\ V_{(r)} 鈭� V_{(鈭�)} = {鈭�}_{畏}^{n} \frac{A}{r^{4}} dr \end{matrix}$

As $V_{鈭�} = 0$ , then

$\begin{matrix} V_{(r)} = 鈭� A {[\frac{r^{3}}{鈭� 3}]}_{r_{1}}^{r_{2}} \\ = \frac{鈭� A}{3} [\frac{1}{r_{2}^{3}} 鈭� \frac{1}{r_{1}^{3}}] \\ = \frac{鈭� A}{3} [\frac{1}{3^{3}} 鈭� \frac{1}{2^{3}}] \\ = + 2.93 脳 10^{鈭� 2} A \end{matrix}$

$V_{(r)} = + A (0.029 / m^{3})$

Hence,the magnitude of the electric potential difference is $+ 2.93 脳 10^{鈭� 2} A$ .

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As a space shuttle moves through the dilute ionized gas of Earth鈥檚 ionosphere, the shuttle鈥檚 potential is typically changed by -1.0 Vduring one revolution. Assuming the shuttle is a sphere of radius 10 m, estimate the amount of charge it collects.

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