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

For the system shown in figure, if the resultant force on q is zero, then \(q=\ldots \ldots \ldots\) (A) \(-2 \sqrt{2} \mathrm{Q}\) (B) \(2 \sqrt{2} \mathrm{Q}\) (C) \(2 \sqrt{3} \mathrm{Q}\) (D) \(-3 \sqrt{2} Q\)

Two identical capacitors have the same capacitance \(\mathrm{C}\). one of them is charged to a potential \(\mathrm{V}_{1}\) and the other to \(\mathrm{V}_{2}\). The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is (A) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}-\mathrm{V}_{2}^{2}\right)\) (B) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}^{2}+\mathrm{V}_{2}^{2}\right)\) (C) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}-\mathrm{V}_{2}\right)^{2}\) (D) \((1 / 4) \mathrm{C}\left(\mathrm{V}_{1}+\mathrm{V}_{2}\right)^{2}\)

The electric potential \(\mathrm{V}\) at any point $\mathrm{x}, \mathrm{y}, \mathrm{z}\( (all in meter) in space is given by \)\mathrm{V}=4 \mathrm{x}^{2}$ volt. The electric field at the point \((1 \mathrm{~m}, 0,2 \mathrm{~m})\) in \(\mathrm{Vm}^{-1}\) is \((\mathrm{A})+8 \mathrm{i} \wedge\) (B) \(-8 \mathrm{i} \wedge\) (C) \(-16 \mathrm{i}\) (D) \(+16 \mathrm{i}\)

Point charges \(q_{1}=2 \mu c\) and \(q_{2}=-1 \mu c\) care kept at points \(\mathrm{x}=0\) and \(\mathrm{x}=6\) respectively. Electrical potential will be zero at points ..... (A) \(\mathrm{x}=-2, \mathrm{x}=2\) (B) \(\mathrm{x}=1, \mathrm{x}=5\) (C) \(\mathrm{x}=4, \mathrm{x}=12\) (D) \(\mathrm{x}=2, \mathrm{x}=9\)

A charge \(\mathrm{Q}\) is divided into two parts and then they are placed at a fixed distance. The force between the two charges is always maximum when the charges are \(\ldots \ldots\) (A) \((Q / 3),(Q / 3)\) (B) \((\mathrm{Q} / 2),(\mathrm{Q} / 2)\) (C) \((Q / 4),(3 Q / 4)\) (D) \((Q / 5),(4 Q / 5)\)
See all solutions

Recommended explanations on English Textbooks

Rhetoric

Read Explanation

Language Analysis

Read Explanation

Pragmatics

Read Explanation

Syntax

Read Explanation

Textual Analysis

Read Explanation

Argumentative Essay

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