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–cubic 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 $V_{(r)} = + 2.93 \times 10^{- 2} A$ .

Step by step solution

Given data:

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

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

Understanding the concept

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

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

Here, $r$ is the distance.

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

Since, the electric field is,

$\begin{matrix} E = \frac{- dV}{dr} \\ \int_{\infty}^{r} dV = - \int_{η}^{n} Edr \\ V_{(r)} - V_{(\infty)} = \int_{η}^{n} \frac{A}{r^{4}} dr \end{matrix}$

As $V_{\infty} = 0$ , then

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

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

Hence,the magnitude of the electric potential difference is $+ 2.93 \times 10^{- 2} A$ .

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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–cubic 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

Step by step solution

Given data:

Understanding the concept

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

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The magnitude E of an electric field depends on the radial distance r according toE=A/r4, where is a constant with the unit volt–cubic meter. As a multiple of A, what is the magnitude of the electric potential difference betweenr=2.00mandr=3.00m?

Short Answer

Step by step solution

Given data:

Understanding the concept

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Engineering Physics

Wave Optics

Translational Dynamics

Radiation

Electric Charge Field and Potential

Study anywhere. Anytime. Across all devices.

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–cubic meter. As a multiple of $A$ , what is the magnitude of the electric potential difference between $r = 2.00 m$ and $r = 3.00 m$ ?