/*! 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 1 A charge \(q\) is accelerated th... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A charge \(q\) is accelerated through a potential difference of \(V\). Find its kinetic energy. a. \(q V\) b. \(q V / 2\) c. \(V\) d. None of these

Short Answer

Expert verified
The kinetic energy is \(qV\), so the answer is \(a.\)

Step by step solution

01

Understanding the Problem

The problem involves a charge \(q\) being accelerated through a potential difference \(V\). We need to find the kinetic energy gained by this charge. The relationship between voltage, charge, and energy is critical here.
02

Applying Energy Principles

When a charge \(q\) moves through a potential difference \(V\), it gains kinetic energy. The energy gained by the charge is equal to the work done, which is calculated by the formula \( W = qV \), where \(W\) is the work done or the kinetic energy gained.
03

Solving the Formula

Substitute the known values into the formula to find the kinetic energy: \( KE = qV \). So the kinetic energy of the charge after being accelerated through the potential difference is \( qV \).
04

Conclusion

Based on the calculations, the kinetic energy of the charge is \( qV \), matching option \(a\). Therefore, the answer is \(a. \, qV\).

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Difference
When we talk about potential difference, we are referring to the difference in electric potential energy per unit charge between two points in space. Often represented by the letter \(V\), it's like the electrical equivalent of mechanical energy's potential energy. Imagine potential difference as an energy booster for charges, giving them the power to move and do work.

Think of it as a charged-up battery. When a charge \(q\) moves from one point to another with a potential difference of \(V\), it experiences a sort of push that transforms its stored potential energy into kinetic energy. This transformation is crucial, as it dictates how the charge will behave and how much energy it gains as it moves.

Potential difference is measured in volts, and it is directly related to the amount of energy transferred to the charge per coulomb. This relationship is fundamental in understanding how electric circuits work and how energy is transmitted and transformed within these circuits.
Energy Principles
Understanding energy principles is key to unraveling how charges interact and move in an electric field. In the realm of electricity, energy principles are centered on the concept of conservation and transformation of energy.

When a charge \(q\) moves through a potential difference \(V\), it undergoes an energy transformation from electrical potential energy to kinetic energy. The principle that guides this transformation is one of the most fundamental laws of physics: the conservation of energy. Essentially, energy cannot be created or destroyed; it can only be transformed from one form to another. In this case, the electrical potential energy is transformed into kinetic energy.

The work-energy principle comes into play here. The work done on the charge is actually the energy that the charge gains, calculated by \(W = qV\). This expression tells us that the amount of kinetic energy gained by the charge is proportional to both the magnitude of the charge and the potential difference through which it moves. This simple yet powerful formula helps us analyze countless electrical situations where energy transformation is involved.
Charge Interaction
Charge interaction is all about the dynamics between different charges and how they influence each other's movement and behavior in an electric field. When charges interact, they exert forces on one another that can result in motion. These interactions help explain how energy is transferred within a system.

The force between charges follows Coulomb's Law, stating that the force experienced by a charge is proportional to the product of the magnitudes of the two charges and inversely proportional to the square of the distance between them. However, when we consider a charge moving through a potential difference, the primary concern is how this affects the kinetic energy of the charge.

As a charge moves through a potential difference, it accelerates, due to the energy gained from the potential difference. The interaction here is crucial as it dictates the transformation of energy states within the charge. By understanding charge interactions, we grasp how electrical energy is used to do work, power electronic devices, and transmit electrical signals efficiently throughout circuits.

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

The electric potential decreases uniformly from \(120 \mathrm{~V}\) to \(80 \mathrm{~V}\) as one moves on the \(X\) -axis from \(x=-1 \mathrm{~cm}\) to \(x=\) \(+1 \mathrm{~cm}\). The electric field at the origin a. must be equal to \(20 \mathrm{~V} \mathrm{~cm}^{-1}\) b. must be equal to \(20 \mathrm{Vm}^{-1}\) c. may be greater than \(20 \mathrm{~V} \mathrm{~cm}^{-1}\) d. may be less than \(20 \mathrm{Vcm}^{-1}\)

Two identical rings \(P\) and \(Q\) of radius \(0.1 \mathrm{~m}\) are mounted coaxially at a distance \(0.5 \mathrm{~m}\) apart. The charges on the two rings are 2 and \(4 \mu \mathrm{C}\), respectively. The work done in transferring a charge of \(5 \mu \mathrm{C}\) from the center of \(P\) to that of \(Q\) is a. \(1.28 \mathrm{~J}\) b. \(0.72 \mathrm{~J}\) c. \(0.144 \mathrm{~J}\) d. \(2.24 \mathrm{~J}\)

A ball of mass \(1 \mathrm{~g}\) carrying a charge \(10^{-8} \mathrm{C}\) moves from a point \(A\) at potential \(600 \mathrm{~V}\) to a point \(B\) at zero potential. The change in its K.E. is a. \(-6 \times 10^{-6} \mathrm{erg}\) b. \(-6 \times 10^{-6} \mathrm{~J}\) c. \(6 \times 10^{-6} \mathrm{~J}\) d. \(6 \times 10^{-6} \mathrm{erg}\)

Four identical charges are placed at the points \((1,0,0)\), \((0,1,0),(-1,0,0)\) and \((0,-1,0)\). Then, a. the potential at the origin is zero b. the electric field at the origin is not zero c. the potential at all points on the \(z\) -axis, other than the origin, is zero d. the field at all points on the \(z\) -axis, other than the origin, acts along the \(z\) -axis

Two conducting spheres of radii \(r_{1}\) and \(r_{2}\) have same elec. tric field near their surfaces. The ratio of their electrical potentials is a. \(\frac{r_{1}^{2}}{r_{2}^{2}}\) b. \(\frac{r_{2}^{2}}{r_{1}^{2}}\) c. \(\frac{r_{1}}{r_{2}}\) d. \(\frac{r_{2}}{r_{1}}\)

See all solutions

Recommended explanations on Physics Textbooks

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.