electric potential problems and solutions pdf

A voltaic cell is constructed based on the oxidation of zinc metal and the reduction of silver cations. endobj Solution: keep in mind that the electric potential is a scalar quantity as opposed to the electric field and force. In this chapter, we'll learn about the electric potential (also called "voltage"), which gives us an alternative, complementary, and useful way to describe electric fields. Creating the right group makeup is also important in ensuring you have the necessary expertise and skillset to both identify and follow up on potential solutions. The minus sign indicates that the potential energy decreases. Substituting the numerical values of the electron into above and solving for $\Delta V$, we will get \begin{align*} \Delta V &=\frac{\frac 12 m(v_f^2-v_i^2)}{e} \\\\ &=\frac{(9.11\times 10^{-31})(0-(8.4\times 10^5)^2)}{2(-1.6\times 10^{-19})} \\\\ &=2 \quad\rm V \end{align*} where the electric charge of electron is $q=-e$. Charge on point A =+9 C and charge on point B = -4 C. How fast does the proton move? Potential Due to Continuous Charge. 2 0 obj The individuals who are preparing for Physics GRE Subject, AP, SAT, ACTexams in physics can make the most of this collection. However if we place a negative charge at P it will move towards the charge +9C. (a) How large is the potential difference between A and B? all the known quantities. (b) How fast does an electron move through the same potential difference? The other processes, electricity transmission, distribution, and electrical power storage and recovery using pumped-storage methods are normally carried out by the electric power industry. Page Published: 2/12/2022. Distribution Problems and Solutions. The magnitude of the electric field between the plates (E) = 500 Volt/meter, The distance between the plates (s) = 2 cm = 0,02 m, The charge on an proton = +1.60 x 10-19 Coulomb, Wanted : The change in electric potential energy (PE). Problem (4): The kinetic energy of a proton is $4.2\,\rm keV$. It was moved to another point Y where its potential energy was 610-2 J. Solution: This type of question appears in all the electric potential problems. To prevent all the potential problems and damages that an uncovered junction can cause, you can get it installed/covered by a professional. (b) The potential difference between $A$ and $B$ is found \[V_B-V_A=-120\,\rm V\] This tells us that point A is at a higher electric potential or $V_A>V_B$. A machine breaks down every 20 days (exponentially distributed). Solution: The work required to move a point charge in the presence of the electric potential of other charges is calculated by \[W=-q\Delta V\] where $\Delta V=V_f-V_i$ is the potential difference created by other charges between the initial and final points. 60. (a) What work was done on the particle by the electric field? In this paper, we summarize the technique of using Green functions to solve electrostatic problems. over 2000 problem and solutions. The first component of the IDEAL approach is to identify potential problems and treat them as opportunities to do something creative. and they satisfy R + T = 1. 19. When in doubt, we can always refer back to the fact that opposite charges attract and like charges repel. There are times when the students can solve the problem but the process of solving is not aligned with the marking scheme of the examination. \begin{align*} W&=q(V_2-V_1) \\ &=(1.6\times 10^{-19})\,(-35-120) \\&=\boxed{-2.48\times 10^{-17}\,\rm J} \end{align*} Electron-volt is another way to measure energy at the microscopic level. 1. This indicates that there must be another work, such as work done by an electric force $W_E$, that has not been given. by electrostatic field, the change in electric potential, V, is. 1. (c) R and S are two charged parallel plates, 0.60 m apart, as shown in Figure 2. But Brazil and its replicators have to exercise great care in design-ing and implementing biofuel programs. global warming), assigning property rights is dicult Coasian solutions are likely to be more eective for small, localized externalities than for larger, more global externalities involving large number of people and rms. The electric charge of proton is $q=+e=1.6\times 10^{-19}\,\rm C$. Customers arrive at the drive-up window at a rate of 12. 2015 All rights reserved. All these questions are for high school and AP Physics exams. If you would like to change your settings or withdraw consent at any time, the link to do so is in our privacy policy accessible from our home page. 3 0 obj On the left, the ball-Earth system gains gravitational potential energy when the ball is higher in Earth's. (b) Compare the potentials at points $A$ and $B$. Problem 1: A particle of mass 40 mg and carrying a charge 510-9 C is moving directly towards fixed positive point charge of magnitude 10-8 C. When it is at a distance of 10 cm from the fixed point charge it has a velocity of 50 cm/s. An electron is to be accelerated in a uniform electric field having a strength of 2.0010 6 V/m . A. b. \[V_B=V_{15}+V_{-8}\] where \begin{align*} V_{15}&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{15\times 10^{-9}}{0.03} \\\\ &=4.5\,\rm V \end{align*} and the potential due to $-8\,\rm nC$ is \begin{align*} V_{-8}&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{-8\times 10^{-9}}{0.03} \\\\ &=-2.4\,\rm V \end{align*} Hence, the potential at point $B$ is \[V_B=4.5+(-2.4)=2.1\,\rm V\] We emphasize again that the sign of electric charges must be included in the formula of electric potential. Application of the method to the problem of Art. Calculate the electric potential associated with the proton's electric field at this distance. Now, Assume the potential is zero, for example, at an arbitrary point at a distance of $x$ from the origin and outside the charges, say $A$. x]o0#?ja{VU'U+.]Ph4H$Lq`SYi4tv^h-0QB (Membranes are discussed in some detail in. The EMC problems are those encountered in EFT, Surge, ESD, power fail and emissions testing. MICAH MORA- Concentration of Solution.pdf. Simplify the block diagram shown in Figure 3-42. Problem (14): In the figure below, two charges are placed at a distance of $d=6\,\rm cm$. Oscillations and Sound Discussions on I E Irodov solutions Problems in General Physics by D B Singh Arihant . Let assume r be the distance between the charges, Consider a small elemental length dx having charge dq, In an electric field say E potential at any point near the dipole is, Electric potential due to electric dipole is given by, Let r be the radius of each small drop and q be the amount of charge on each one of them. Exam 1 Practice Problems Solutions. (mol), and luminosity in candela (cd). Line-integrals and potentials 17. After the original proposal for the mechanism of K+ transport in yeast [1], it was shown [2, 3] that this ion is transported because a H+-ATPase exists in the plasma membrane, pumping protons outside, therein generating an electric membrane potential difference (PMP), negative inside. 19. Electric polarization and displacement 61. Thus, the potential difference between initial and final points is \begin{align*} V_f-V_i&=\frac{W}{-q} \\\\ &= \frac{1.8\times 10^{-6}}{-3.6\times 10^{-9}} \\\\ &=-500\,\rm V \\\\ \Rightarrow \quad \boxed{V_f=V_i-500 }\end{align*}The above statement tells us that the end potential is $500\,\rm V$ less than the start potential, as expected. Ferghana, Uzbekistan. How much work was done by the electric fields due to those two charges on a charge of $1.5\,\rm nC$ that moves from point $B$ to point $A$? By definition, the difference between the final and initial potentials, called potential difference $\Delta V$, is \begin{gather*} \Delta V=V_f-V_i \\ -1.875 =V_f-V_i \\ \Rightarrow \quad \boxed{V_f=V_i-1.875} \end{gather*} Therefore, the potential at a point $2\,\rm m$ away from the origin is lower than the potential at the origin. Two parallel plates are charged. Solution. Now, let's take a look at the most common electrical problems and solutions! (a) Substitute the numerical values related to the mass and charge of the proton into the above formula \begin{align*} v&=\sqrt{\frac{2q\Delta V}{m}} \\\\&=\sqrt{\frac{2(1.6\times 10^{-19})(120)}{1.67\times 10^{-27}}} \\\\&= 152\times 10^{3}\,\rm m/s \end{align*} Problem 4 Determine the length of a flat-belt pulley drive having the following data: Diameter of first Example problems and solutions A-3-1. So I encourage anyone interested in environmental solutions to think big-picture. electric potential, the amount of work needed to move a unit charge from a reference point to a specific point against an electric field. Manage SettingsContinue with Recommended Cookies. Section 25.8-Q.75) (a) A uniformly charged cylindrical shell has total charge Q, radius R, and height h. Determine the electric potential at a point a distance d from the right end of the cylinder, as shown in Figure P25.75. Question 1 Calculate the amount of work done in assembling charge together to form a uniformly charged sphere. endobj (b) point $A$ as shown. ABSTRACT The following is the very rst set of the series in 'Problems and Solutions in a Graduate Course in Classical Electrodynamics'. Part I: Short Questions and Concept Questions Problem 1: Spark Plug Pictured at right is a typical spark Problem 2 Consider the three charges +Q , +2Q , and !Q , and a mathematical spherical surface (it does not physically exist) as shown in the figure below. Electrical problems can happen anywhere where electricity is. Lesson_2_Electric_Potential_and_Electric_Potential_Energy.pdf. Abstract This paper will discuss actual EMC problems encountered in electronic products. Since nodes 1 and 2 must be at the same potential, there is no potential difference across R 1 . 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What is the electric potential at (a) a point in the middle of the two charges? An electric charge of 210-3 C at a point X in an electric field had electric potential energy of 410-2 J. How much work would be required to displace a test point charge of $0.2\,\rm \mu C$ from a point midway between them to a point $12\,\rm cm$ closer to either of the charges? Also, it is the work that needs to be done to move a unit charge from a reference point to a precise point inside the field with production acceleration. Manage SettingsContinue with Recommended Cookies. By using an energy approach, problems could be solved that were insoluble using forces. Find the electric potential at the same location. Electrons are free to move in a conductor. This paper will show how to correct these problems using various methods. 5 0 obj Substituting the numerical values into the above expression, we have \begin{gather*}-q(V_f-V_i)=\frac 12 m(v_f^2-v_i^2) \\\\ -(-3.6) \Delta V=\frac{0.045\times \left(10^2-20^2\right)}{2} \\\\ \Rightarrow \quad \Delta V=\frac{-6.75}{3.6}=-1.875\,\rm V \end{gather*} we can use this potential difference and determine which point is at higher or lower potential. In this question, the electric potential at two different points is given, and asked the amount of work done on the proton with a charge of $q=+1.6\times 10^{-19}\,\rm C$. The potential difference between the points P and Q is given by. But there is no magic bullet for environmental problems. Problem (17): In the following configuration, what is the electric potential difference $V_A-V_B$? Solutions: problem set 2. Solution: The magnitude of the electric potential difference $\Delta V$ and the electric field strength $E$ are related together by the formula $\Delta V=Ed$ where $d$ is the distance between the initial and final points. "In thermodynamics, chemical potential, also known as partial molar free energy (wrong), is a form of potential energy (wrong) that can be. r <0 Work is positive Positive Work -->Energy is taken from the Stored U and given to the object (KE) Negative Work --> Energy is stored in U 2 Electric Potential Energy Multiple Charges 1 qq qq qq U = k 1 2 + k 2 3 +k 3 1 r31 r12 r23 r12 2. (To stop an electron, a negative potential difference must be applied.) Hence, no current flows through R 1 . According to the Center for Environmental Solutions, almost 97 % of Belarusians have medicines. <>>> The separation between the plates is 2 cm and the magnitude of the electric field between the plates is 500 Volt/meter. The potential due to the charge $4\,\rm \mu C$ at point $A$ is \begin{align*} V_4&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{4\times 10^{-6}}{0.03} \\\\ &=1.2\times 10^6\,\rm V \end{align*} For the charge $1\,\rm \mu C$, we have \begin{align*} V_1&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{1\times 10^{-6}}{0.02} \\\\ &=0.45\times 10^6\,\rm V \end{align*} And similarly, for the charge $-2\,\rm \mu C$, \begin{align*} V_{-2}&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{-2\times 10^{-6}}{0.05} \\\\ &=-0.36\times 10^6\,\rm V \end{align*} Note that to find the distance of the charge $-2\,\rm \mu C$ from the point $A$, we applied the Pythagorean theorem as below \[\sqrt{2^2+3^2}=5\,\rm cm\] Next, we simply add all these potentials together. The repair service with which the Laundromat contracts takes an average, The First American Bank of Rapid City has one outside drive-up teller. This question was similar to the work-kinetic energy problems.if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[250,250],'physexams_com-large-mobile-banner-1','ezslot_2',154,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-large-mobile-banner-1-0'); Problem (7): A $+9\,\rm \mu C$ charge moves from the origin to a point of coordinate $(x=2\,\rm cm,y=5\,\rm cm)$ in a uniform electric field of magnitude $240\,\rm V/m$. This page contains answers to "CES test for electrical engineers, electronic and control engineering", and serve as a database of questions and answers, using which seafarer can prepare to exams for getting certificate of competence, or just to challenge yourself knowledge in this theme. Combining these two equations and solving for $\Delta V$, we get \begin{gather*} K_f-K_i=W_F-q\Delta V \\\\ \Rightarrow \Delta V= \frac{K_f-K_i-W_F}{-q} \\\\ = \frac{(3.5\times 10^{-4})-(9.5\times 10^{-4})}{-(-5\times 10^{-6})} \\\\ =\boxed{-120\quad \rm V} \end{gather*} where we set the initial kinetic energy $K_i=0$, since the charge is initially at rest, $v_i=0$. Which point is at the lower potential? Detailed step-by-step solutions are provided for all problems. Problem (3): Suppose an electron moving at a constant speed of $8.4\times 10^5\,\rm m/s$. Also, you can collect a number of potentially difficult lexical items, put them in a column on the board, and ask students to read them out one by one. In addition, there are hundreds of problems with detailed solutions on various physics topics. An electron is accelerated from rest through a potential difference 12 V. What is the change in electric potential energy of the electron? $.' We then provide a variety of example problems in spherical, Cartesian, and cylindrical coordinates. Problem (1): How much work does the electric field do in displacing a proton from a position at a potential of $+120\,\rm V$ to a point at $-35\,\rm V$? (i) On Figure 2, sketch the electric field between R and S, showing its direction. % Two point charges are separated by a distance of 10 cm. The electric potential at an arbitrary point at a distance $r$ from a point charge is $V=k\frac{q}{r}$. Description: This book is the companion guide to Electrical Engineering: License Review. The ions are produced by introducing the sample into an ICP which strips off electrons thereby creating positively charged ions. (a) In this part, we can simply use the work-kinetic energy theorem, $\Delta K=W$, and find the work done by the electric force. To solve such problems, keep in mind that, you must first find the kinetic energy $K=\frac 12 mv^2$ of the accelerated point charge through the given potential difference. (b) Similarly, the potential at point $A$ is \[V_A=V_{15}+V_{-8}\] where \begin{align*} V_{15}&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{15\times 10^{-9}}{0.06} \\\\ &=2.25\,\rm V \end{align*} and \begin{align*} V_{-8}&=k\frac{q}{r}\\\\ &=(9\times 10^9)\frac{-8\times 10^{-9}}{0.06} \\\\ &=-1.2\,\rm V \end{align*} Therefore, we have \[V_A=2.25+(-1.2)=1.05\,\rm V\]. Problem 1 Problem 6: What distance must separate two charges of + 5.610 -4 C and -6.310 -4 C in order to have an electric potential energy with a magnitude of 5.0 J in the system of the two charges? SI units for electromagnetic quantities such as coulombs (C) for charge and volts (V) for electric potential are derived from these fundamental units. Problem (5): The potential difference between two parallel plates $7.5\,\rm mm$ apart is $240\,\rm V$. What energy in keV is given to the electron if it is accelerated through 0.400m?______keV. The consent submitted will only be used for data processing originating from this website. Solution: We can see this as a work problem in physics that is done by some forces. Figure 3-62 Operational-amplifier circuit. Standard Reduction Potentials (volts) in Aqueous Solution. 164. Question 4 Two circular wire loops of radii .09 m (loop I) and .05 m(Loop II) are placed such that their axes coincide and their center are .12 m apart. 55 56 57. What is the potential difference between the origin and that point? In each of the sets of the problems we intend to follow a theme, which not only makes it unique but also deals with the investigation of a certain class of problems in a. To view the purposes they believe they have legitimate interest for, or to object to this data processing use the vendor list link below. If there was a potential difference between two points, then an electric field must exist. Thus, a positive charge will be attracted by a negative poten-tial, and hence ow toward it, and vice versa: electrons, which have negative charge, ow toward a positive potential or voltage. The charge initially is at rest and finally acquires kinetic energy of $3.5\times 10^{-4}\,\rm J$ at point B. (b) the electric field, and (c) electric potential energy. the XXI international Symposium "Dynamic and technological problems of mechanics of structures and continuous media" named after A. G. Gorshkov. k = 9 x 109 Nm2C2, 1 C = 106 C. What is the change in electric potential energy of charge on point B if accelerated to point A ? 38. Use the electron volt to express energy and solve simple problems applying energy conservation. The motion of electricity analogous to that of an incompressible. Course Hero is not sponsored or endorsed by any college or university. 9.4 Electrostatic Potential Energy 9.5 Summary 9.6 Terminal Questions 9.7 Solutions and Answers. Problem 3.2 Again, consider a particle incident on an infinite planar surface sep-arating empty space and an infinite region with constant potential energy V. Now,though, take the energy of the particle to be E < V (Fig. Solutions of silver nitrate and zinc nitrate also were used. Some of our partners may process your data as a part of their legitimate business interest without asking for consent. (ii) Its magnitude is given by the change in the magnitude of potential. They can be done in groups and pairs, in a funny or a bored voice, quietly or loudly, you name it. Its time period is found to be T. Now the space between the plates is made gravity free, and an electric a. An electron is accelerated from rest through a potential difference 12 V. What is the change in, 2. 1. Intravenous therapy. 2.9 Conductivity and mobility 2.10 Liquid junction potentials 2.11 Liquid junction potentials, ion-selective electrodes, and. What electric field strength does form between them? if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[250,250],'physexams_com-narrow-sky-1','ezslot_13',134,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-narrow-sky-1-0'); In this case, the initial point is in the middle of two identical charges. Carefully consider who to include at each stage to help ensure your problem-solving method is followed and positioned for success. stream The magnitude of decrease in the kinetic energy is equal the work done by the electric forces or \[\Delta K=\underbrace{q\Delta V}_{W} \] where $\Delta K=\frac 12 m(v_2-v_1)^2$. You can find the free PDF of HC Verma Solutions for Class 12 Physics (Part 2) for Chapter 29 Electric Field and Potential on Vedantu's site. To solve such problems, keep in mind that, you must first find the kinetic energy $K=\frac 12 mv^2$ of the accelerated point charge through the given potential difference. 5 Electric field outside: (2) Outside the cloud =0 =-0 b Laplace's equation Electric field outside: Electric field continuity at R=b. Including multiple parts, there are 600 problems in the text and solutions are presented here for the majority of them. Suppose at any instant r be the radius of the sphere and now we add a charged shell of radius dr to this sphere of radius r. This process continues till the radius of sphere becomes equal to R. Now, charge on sphere of radius r having volume charge density $\rho$ is. Adjacent points that have the same electric potential form an equipotential surface, which can be either an imaginary surface or a real, physical surface. We can define an electrostatic potential energy, analogous to gravitational potential energy, and apply the law of conservation of energy in the analysis of electrical problems. Electric potential and electric potenial energy. The separation between the plates is 2 cm and the magnitude of the, 3. If you would like to change your settings or withdraw consent at any time, the link to do so is in our privacy policy accessible from our home page. Geometrical theorems about inversion. We can easily understand the concept of electric potential energy by comparing this with gravitational potential energy. Power quality is a set of electrical boundaries that allows a piece of equipment to function in its intended manner without significant loss of performance or life expectancy. \[V_f 1. Its kinetic energy is measured to be $1.8\times 10^{-6}\,\rm J$ after traveling a distance of $4\,\rm cm$. Problem (6): To move a charge of $-5\,\rm \mu C$ from point A to point B, a work of $9.5\times 10^{-4}\,\rm J$ is done by an unknown external force. Q.24-1. We assume in a region away from the edges of the two parallel plates, the electric field is uniform. What is the change in electric potential energy of charge on point B if accelerated to point A ? Cand. The new time c. 32.5 x 10-8 s period. You may assume a uniform electric field. Two point charges of +2.5 C and -6.8 C are separated by a distance of 4.0 m. What is the electric potential midway between the charges? Through what electric potential difference did the charge move? Solution: The electric potential due to a point charge $q$ at distance $r$ from that is calculated by the formula \[V=k\frac{q}{r}\] where $k=8.99\times 10^9 \,\rm \frac{N\cdot m^2}{C^2}$ is the coulomb's constant. Over what distance would it have to be accelerated to increase its energy by 50 GeV? Moreover, over in this topic, we will learn the electric potential, electric potential formula, formula's derivation, and solved example. 8. (ii) Point T is mid-way between R and S. Calculate the electric field strength at T. All remaining questions refer to the following scenario.- Suppose we have two 1 metre long rods. (easy) A negatively charged particle (q = -2 C) moves through a 2000 v loss of electric potential. To view the purposes they believe they have legitimate interest for, or to object to this data processing use the vendor list link below. \[\Delta K=W_F+W_E\] On the other hand, we know that the work done by an electric force equals $W_E=-q\Delta V$. (b) To find the speed of an electron accelerated from rest through a potential difference $\Delta V$, put the numerical values of an electron into the above expression \begin{align*} v&=\sqrt{\frac{2q\Delta V}{m}} \\\\&=\sqrt{\frac{2(1.6\times 10^{-19})(120)}{9.11\times 10^{-31}}} \\\\&= 6.42\times 10^{6}\,\rm m/s \end{align*}. Solution: According to the work-kinetic energy theorem, the work done on an object between two points by some forces could cause that object to gain kinetic energy \[\Delta K=W\] (a) In this problem, the work done by the external force is not equal the kinetic energy of the charge at point B. Use these results and symmetry to find the potential at as many points as possible without additional calculation. In the previous chapter we learned about the use of the electric field concept to describe electric forces between charges. 2. Suppose, using an xyz coordinate system, in some region of space, we find the electric potential is. Solution: This type of question appears in all the electric potential problems. What you'll learn: Electric Potential Energy Problems Definition. solution and moved around until a. point is found where no lOOO-cps. Find the electric potential at the five points indicated with open circles. <> 1.1. Also includes a complete 8-hour practice exam. Introduction Given the limits of fossil and nuclear resources and the social. As a result, that point is placed $12\,\rm cm$ from, say right charge, and $50-12=38\,\rm cm$ from the left charge as shown in the figure. . (3) Solve for the unknown concentration and use that. 15. <> The environmental and social risks of biofuel development, also demonstrated in Brazil, are great and could well undermine all of the potential advantages if not. 4 0 obj Express your result both in joules and electron-volt. Cars arrive at the regular unleaded pump at a rate of 6 per hour. In this case, the charge travels from point $B$ to point $A$, so we must first find the potential difference between these two points. Problem (8): A particle having charge $q=-3.6\,\rm \mu C$ and mass $m=0.045\,\rm kg$ has an initial velocity $v_i=20\,\rm m/s$ at the origin. Potential at the surface of each drop is, Class 9 Science Chapter 10 Gravitation online Test, Online Test for Class 11 Physics Mechanical Properties of Fluids, Class 9 Maths Chapter -3 Coordinate Geometry MCQs, Relation between electric fields and electric potential, Potential energy of dipole placed in uniform electric field, Synthetic Fibres and Plastics Class 8 Practice questions, Class 8 science chapter 5 extra questions and Answers. The change in electric potential energy : 3. \[V_{tot}=V_2+V_2=2\times 22475=44950\,\rm V\]. The charge on an electron (e) = -1.60 x 10-19 Coulomb, Electric potential = voltage (V) = 12 Volt, Wanted: The change in electric potential energy of the electron (PE), PE = q V = (-1.60 x 10-19 C)(12 V) = -19.2 x 10-19 Joule. The potential at point $A$ is $V_A=V_1+V_2$, where \begin{align*} V_1&=k\frac{q}{r} \\\\&=(9\times 10^9) \frac{2.6\times 10^{-9}}{0.13} \\\\&=180\,\rm V \end{align*} and \begin{align*} V_2&=k\frac{q}{r} \\\\&=(9\times 10^9) \frac{-4.6\times 10^{-9}}{0.05} \\\\&=-828\,\rm V \end{align*} Therefore, the potential at point $A$ is \[\boxed{V_A=180-828=-648\,\rm V}\] Similarly, the potential at point $B$ is $V_B=V_1+V_2$, where \begin{align*} V_1&=k\frac{q}{r} \\\\&=(9\times 10^9) \frac{2.6\times 10^{-9}}{0.06}\\\\&=390\,\rm V \end{align*} and \begin{align*} V_2&=k\frac{q}{r} \\\\&=(9\times 10^9) \frac{-4.6\times 10^{-9}}{0.06}\\\\&=-690\,\rm V \end{align*} Thus, the potential at point $B$ is \[\boxed{V_B=390-690=-300\,\rm V}\] The $1.5-\,\rm nC$ charge moves from $B$ to $A$, so the potential difference between points these points is \begin{align*} \Delta V&=V_{final}-V_{initial} \\&=V_A-V_B \\&=-648-(-300) \\&=\boxed{-148\,\rm V}\end{align*} By having the potential difference between the two points, the work done can be easily obtained as follows \begin{align*} W&=-q\Delta V \\ &=-(1.5\times 10^{-9})(-148) \\&=0.22\times 10^{-6}\,\rm J \end{align*}, Author: Dr. Ali Nemati When problems are treated as opportunities, the result is often a solution or invention that otherwise would have eluded you. R0 > R (i.e. if(typeof ez_ad_units != 'undefined'){ez_ad_units.push([[300,250],'physexams_com-box-4','ezslot_3',114,'0','0'])};__ez_fad_position('div-gpt-ad-physexams_com-box-4-0'); Solution: The magnitude of the electric potential difference between two points in a uniform electric field $E$ is found by \[\Delta V=Ed\] where $d$ is the distance between the two points. Some of our partners may process your data as a part of their legitimate business interest without asking for consent. Finding fields from potentials - Determine the electric field when given an electric potential that is a function of one position variable only. The consent submitted will only be used for data processing originating from this website. (a) How fast does the proton move across this region? 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