/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 The electric potential \(V\) (in... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 65

The electric potential \(V\) (in volt) varies with \(x\) (in metre) according to the relation \(V=5+4 x^{2}\). The force experienced by a negative charge of \(2 \times 10^{-6} \mathrm{C}\) located at \(x=0.5 \mathrm{~m}\) is (A) \(2 \times 10^{-6} \mathrm{~N}\) (B) \(4 \times 10^{-6} \mathrm{~N}\) (C) \(6 \times 10^{-6} \mathrm{~N}\) (D) \(8 \times 10^{-6} \mathrm{~N}\)

Short Answer

Expert verified
The force experienced by the negative charge at x = 0.5m is \(8 \times 10^{-6} N\), which corresponds to option (D).

Step by step solution

01

Find the electric field (E(x)) from the potential (V(x))

To find the electric field (E(x)), we'll need to calculate the derivative of the potential with respect to the position (x). The formula for this is: \[ E(x) = - \frac{dV(x)}{dx} \] So first, let's find the derivative of V(x): \[ V(x) = 5 + 4x^2 \] \[ \frac{dV(x)}{dx} = 8x \] Now, calculate the electric field (E(x)): \[ E(x) = - \frac{dV(x)}{dx} = -8x \]
02

Calculate the electric field at x = 0.5m

Now that we have the electric field as a function of position, we can find the electric field at the specific location (x = 0.5m) where the charge is located. \[ E(0.5) = -8(0.5) = -4 \, V/m \]
03

Calculate the force on the negative charge

Finally, we can find the force experienced by the negative charge using the formula F = qE, where F is the force, q is the charge (2 x 10^{-6} C), and E is the electric field at the location of the charge. \[ F = qE = (2 \times 10^{-6} \, C)(-4 \, V/m) \] The negative sign indicates the force is in the opposite direction of the electric field. Since we only need the magnitude of the force, we can ignore the negative sign: \[ F = - (2 \times 10^{-6} \, C)(-4 \, V/m) = 8 \times 10^{-6} \, N \] So the force experienced by the negative charge at x = 0.5m is \(8 \times 10^{-6} N\), which corresponds to option (D).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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 spring block system undergoes vertical oscillations above a large horizontal metal sheet with uniform positive charge. The time period of the oscillation is \(T\). If the block is given a charge \(Q\), its time period of oscillation (A) remains same. (B) increases. (C) decreases. (D) increases if \(Q\) is positive and decreases if \(Q\) is negative.

Two equal positive charges \(+q\) each are fixed a certain distance apart. A third equal positive charge \(+q\) is placed exactly midway between them. Then the third charge will (A) move at an angle of \(45^{\circ}\) to the line joining the two charges. (B) move at an angle of \(90^{\circ}\) to the line joining the two charges. (C) move along the line joining the two charges. (D) stay at rest.

Assertion: When an uncharged capacitor of capacitance \(C\) is charged by a cell of emf \(V\), the energy stored by capacitor is \(\frac{1}{2} C V^{2}\), and energy supplied by battery is \(C V^{2}\). Reason: In charging an uncharged capacitor, energy is lost in the form of heat. (A) A (B) \(\mathrm{B}\) (C) \(\mathrm{C}\) (D) \(\mathrm{D}\)

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

Two point charges \(+4 q\) and \(+q\) are placed \(30 \mathrm{~cm}\) apart. At what point on the line joining them is the electric field zero? (A) \(15 \mathrm{~cm}\) from charge \(4 q\) (B) \(20 \mathrm{~cm}\) from charge \(4 q\) (C) \(7.5 \mathrm{~cm}\) from charge \(q\) (D) \(5 \mathrm{~cm}\) from charge \(q\)
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Oscillations

Read Explanation

Translational Dynamics

Read Explanation

Rotational Dynamics

Read Explanation

Modern Physics

Read Explanation

Conservation of Energy and Momentum

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.