Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 1639

The potential at a point \(\mathrm{x}\) (measured in \(\mu \mathrm{m}\) ) due to some charges situated on the \(\mathrm{x}\) -axis is given by \(\mathrm{V}(\mathrm{x})=\left[(20) /\left(\mathrm{x}^{2}-4\right)\right]\) Volt. The electric field at \(\mathrm{x}=4 \mu \mathrm{m}\) is given by (A) \((5 / 3) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (B) \((10 / 9) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (C) \((10 / 9) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction (D) \((5 / 3) \mathrm{V} \mu \mathrm{m}^{-1}\) and in positive \(\mathrm{x}\) - direction

Short Answer

Expert verified
The electric field at \(x = 4 \mu m\) is \(\frac{10}{9} V\mu m^{-1}\) and is found to be in positive x-direction. Hence, the correct answer is (C).

Step by step solution

01

Compute the derivative of V(x) with respect to x

To calculate the derivative of V(x), we can use the power rule. We get: \[\frac{dV}{dx} = \frac{d}{dx}[\frac{20}{x^2-4}]\] First, we rewrite V(x) as a power of x: \(20(x^2-4)^{-1}\). Then, applying the power rule: \[\frac{dV}{dx}=-20(-1)(x^2-4)^{-2}(2x)\] Simplifying, we get: \[\frac{dV}{dx}=\frac{40x}{(x^2-4)^2}\]
02

Find the electric field E(x) at \(x = 4 \mu m\)

Now we have the derivative of V(x), which gives us the electric field E(x): \[E(x)=\frac{40x}{(x^2-4)^2}\] We want to calculate the electric field at \(x = 4 \mu m\). So we substitute the given value: \[E(4) = \frac{40(4)}{(4^2-4)^2}\]
03

Determine the direction of the electric field

Since E(x) has a positive value for \(x = 4 \mu m\), the electric field's direction points towards the positive x-direction. Now we compute the value of E(4): \[E(4) = \frac{40(4)}{(4^2-4)^2} = \frac{40(4)}{(16-4)^2} = \frac{40(4)}{12^2} = \frac{10}{9} V\mu m^{-1}\] The electric field at \(x = 4 \mu m\) is \(\frac{10}{9} V\mu m^{-1}\) and is found to be in positive x-direction. Hence, the correct answer is (C).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A parallel plate capacitor with air between the plates has a capacitance of $9 \mathrm{pF}\(. The separation between its plates is \)\mathrm{d}$. The space between the plates is now filled with two dielectrics. One of the dielectric constant \(\mathrm{K}_{1}=3\) and thickness \(\mathrm{d} / 3\) while the other one has dielectric constant \(\mathrm{K}_{2}=6\) and thickness \(2 \mathrm{~d} / 3\). Capacitance of the capacitor is now (A) \(1.8 \mathrm{pF}\) (B) \(20.25 \mathrm{pF}\) (C) \(40.5 \mathrm{pF}\) (D) \(45 \mathrm{pF}\)

An electric dipole is placed at an angle of \(60^{\circ}\) with an electric field of intensity \(10^{5} \mathrm{NC}^{-1}\). It experiences a torque equal to \(8 \sqrt{3} \mathrm{Nm}\). If the dipole length is \(2 \mathrm{~cm}\) then the charge on the dipole is \(\ldots \ldots \ldots\) c. (A) \(-8 \times 10^{3}\) (B) \(8.54 \times 10^{-4}\) (C) \(8 \times 10^{-3}\) (D) \(0.85 \times 10^{-6}\)

At what angle \(\theta\) a point \(P\) must be located from dipole axis so that the electric field intensity at the point is perpendicular to the dipole axis? (A) \(\tan ^{-1}(1 / \sqrt{2})\) (B) \(\tan ^{-1}(1 / 2)\) (C) \(\tan ^{-1}(2)\) (C) \(\tan ^{-1}(\sqrt{2})\)

A thin spherical shell of radius \(R\) has charge \(Q\) spread uniformly over its surface. Which of the following graphs, figure most closely represents the electric field \(\mathrm{E}\) (r) produced by the shell in the range $0 \leq \mathrm{r}<\infty\(, where \)\mathrm{r}$ is the distance from the centre of the shel1.

Electric potential at any point is $\mathrm{V}=-5 \mathrm{x}+3 \mathrm{y}+\sqrt{(15 \mathrm{z})}$, then the magnitude of the electric field is \(\ldots \ldots \ldots \mathrm{N} / \mathrm{C}\). (A) \(3 \sqrt{2}\) (B) \(4 \sqrt{2}\) (C) 7 (D) \(5 \sqrt{2}\)
See all solutions

Recommended explanations on English Textbooks

Blog

Read Explanation

Lexis and Semantics Summary

Read Explanation

Rhetoric

Read Explanation

Summary Text

Read Explanation

Sociolinguistics

Read Explanation

Cues and Conventions

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free