Lecture 24: Electric Potential PDF

Lecture_24 Electric Potential Let's start understanding the forces between charges and how they relate to electric potential energy. The Basics of Electric Charges and Forces In electrostatics, we deal with charges that can be either positive or negative. Wh...

Lecture_24 Electric Potential Let's start understanding the forces between charges and how they relate to electric potential energy. The Basics of Electric Charges and Forces In electrostatics, we deal with charges that can be either positive or negative. When there are two positive or two negative charges, a force arises between them. This force is repulsive, meaning the charges push away from each other. Conversely, a positive and a negative charge will attract each other. The force between these charges is crucial in understanding electric potential energy. Work and Potential Energy When we bring one charge close to another, we must exert a force to overcome the repulsive, as shown in fig. 24.1, or attractive force between them. This exertion of force is called work. As we do work to move a charge, potential energy is added to the system. This potential energy is what gives rise to the concept of electric potential. Figure 24.1: Two opposite charges repel each other. To illustrate this, imagine pressing a spring. Pressing the spring requires work against the force it exerts, storing potential energy in the compressed spring. Similarly, when we bring two charges close together, we do work against the electric force, thereby increasing the electrical potential energy. Visualizing Electric Potential Energy Consider two charges being brought close to each other. A simple animation, given in ppt file, can help visualize this process. Imagine a monkey holding one charge and bringing it close to another fixed charge. The monkey must do work to bring the charges closer, but if the monkey releases the charge, all the work done is recovered as kinetic energy. This demonstrates that no energy is lost, and all the energy stored in the system can be recovered. Calculating Electric Potential Energy Let's imagine one charge is fixed in place, and we want to bring another charge, a unit charge (q = 1), from a far distance to a specific distance R. We need to calculate the total work required for this process. This calculation is divided into two parts.  1. Starting from Infinity: We begin by considering the unit charge at infinity, where the electric potential energy is considered zero. 2. Moving to Distance R: We move the charge to a distance R, calculating the work done step by step. The work done in moving a unit charge from infinity to a distance R is given by the integral of the force over the distance: R W =  Edr  Here, E represents the electric field, and dr is the infinitesimal distance. The electric field E is derived from Coulomb's law: 1 q E= 4 0 r 2 Substituting this into our integral: R 1 dr W= q 4 0  r2 This integral simplifies as: R 1 W= q  r −2 dr 4 0  R 1  r −2+1  W= q  4 0  −2 + 1   R 1  r −1  W= q  4 0  −1   R 1  1 W= q −  4 0  r   Evaluating this from infinity to R: q 1 1 W=  −  4 0  R   q 1 1 W= =0 4 0 R  Potential Energy Between Two Charges If we have two charges, q1 and q2, the potential energy U between them when separated by a distance r is:  1 q1q2 U (r ) = 4 0 r We assume the potential energy is zero when the charges are infinitely far apart. This simplifies our calculations and provides a reference point. Units of Electric Potential The unit of electric potential is the volt (V), which is equivalent to one joule per coulomb (J/C). When you multiply the electric potential (in volts) by the charge (in coulombs), you get the energy (in joules). Electron Volt A useful unit in atomic and nuclear physics is the electron volt (eV). One electron volt is the energy gained or lost by an electron when it moves through an electric potential difference of one volt. This is equal to 1.6×10−19 joules. Applications of Electric Potential In atomic physics, the electron volt is particularly significant. For instance, to ionize a hydrogen atom (i.e., to remove an electron from a proton), you need to provide 13.6 electron volts of energy. In nuclear physics, the binding energy of protons and neutrons in a nucleus is often measured in million electron volts (MeV). Movement of Charges in Electric Potential A positive charge naturally moves from a region of higher potential to a region of lower potential, whereas a negative charge moves in the opposite direction. This behaviour can be visualized with a simple experiment. If you place a charge in an electric field, the charge will move according to the potential difference between two points. Work Done by Electric Field Consider a charge moving from point A to point B in an electric field. The work done by the electric field on the charge can be calculated. For instance, if the electric field does 8 joules of work on a charge of 2 coulombs moving from point A to point B, as shown in fig. 24.2, the potential difference is: W V = − q 8 V = − = −4V 2 The negative sign indicates a decrease in potential. Figure 24.2: Change in potential as charge moves from A to B. Conservative and Non-Conservative Forces The force between two charges is conservative, meaning the work done in moving a charge between two points is path-independent. This property allows us to define potential energy and potential differences accurately. In contrast, non-conservative forces, such as friction, depend on the path taken. Electrostatic Potential Energy in Multi-Charge Systems Understanding the electrostatic potential energy in systems of multiple point charges is fundamental in electrostatics. Consider a system of three-point charges q1, q2, and q3, as shown in fig. 24.3. The total electrostatic potential energy U(r) of this system is given by 1 q1q2 1 q1q3 1 q2 q3 U (r ) = + + 4 0 r12 4 0 r13 4 0 r23 1  q1q2 q1q3 q2 q3  U (r ) =  + +  4 0  r12 r13 r23  Figure 24.3: Electrostatic potential energy for three charges. Dipole A dipole consists of two equal but opposite charges separated by a distance d, as shown in fig. 24.4. The dipole moment p is defined as: P=q⋅d Dipoles are significant in many molecules where there is a natural separation of charge, such as in water (H2O). Figure 24.4: A dipole separated by distance d. To calculate the potential V at a point P due to a dipole, consider the distances from the charges to the point P, as shown in fig. 24.5: Figure 24.5: Potential at point P due to a dipole.  VP = V1 + V2 1  q −q  =  +  4 0  r1 r2  q r2 − r1 = 4 0 r1r2 For points far from the dipole, where r is the distance from the centre of the dipole to the point P and θ is the angle between the dipole axis and the line joining the centre of the dipole to the point P: For r  d r2 − r1  d cos  r1r2  r 2 q d cos  1 p cos  V = p=q.d 4 0 r 2 4 0 r 2  Note that V = 0 at  = 2 Ring of Charges Consider a uniformly charged ring lies in xy-plan with its center at origin, as shown in fig. 24.6, have a total charge q and radius R. To find the potential at a point P along the axis of the ring, use the symmetry of the problem. The potential due to a point charge dq at a distance r is given by: 1 dq dV = eq(1) 4 0 r For a small element of charge dq on the ring, at coordinates (R cos , R sin  ,0) , the distance to point P is: r = R 2 + z 2 eq(2) Since the ring is uniformly charged, the charge density λ (charge per unit length) is: Figure 24.6: Potential due to a q charged ring at perpendicular axis. = 2 R An infinitesimal charge element is: q q dq =  Rd = Rd = d eq(3) 2 R 2 Integrate dV over the entire ring: V =  dV using eq(1) and eq(2), we get 2 1 dq V = 0 4 0 R2 + z 2 using eq(3), we get q d V = 2 1 2 0 4 0 R2 + z 2 1 q 2 V= 4 0 2 R 2 + z 2 0  d 1 q 2 V= 4 0 2 R 2 + z 2 2  0 d = 2 q V= 4 0 R2 + z 2 So, the potential at a point P on the axis of the ring is: 1 q V= 4 0 R2 + z 2 Disk of Charges For a uniformly charged disk lies in the xy-plane with its centre at the origin have surface charge density  and radius R, as shown in fig. 24.7, we want to find the electric potential V at a point P on the axis perpendicular to the disk at a distance z from the centre of the disk. Consider an infinitesimal ring of radius  and thickness d within the disk. The charge dq on this ring is: dq =   2  d The distance from any point on this ring to point P is: r = 2 + z2 The potential dV due to this ring at point P is: Figure 24.7: Potential due to a disk of uniform surface charge density at perpendicular axis.  1 dq dV = 4 2 + z2 0 1   2  d  dV = 4 0 2 + z2   d dV = 2 0 2 + z2 R   d V = 0 2 0 2 + z2 du let u =  2 + z 2 then ( du = 2 d  ) and  d = 2  R2 + z 2 du V= 2 0  z 2 2 u    ( ) R 2 + z2 V= 2 u  = R2 + z 2 − z 4  z 0 2 2 0 So, the potential at point P on the axis perpendicular to the disk is: V=  2 0 ( R2 + z 2 − z ) Understanding Electric Potential: Imagine a uniform electric field. This means the strength of the electric field doesn't change as you move through it. Now, consider a scenario where we have two points, a and a, in this field. If we move a charge (let's call it b) from point b to point a, as shown in fig. 24.8, we have to do work against the electric field. This work done by the electric field is the force pushing the charge multiplied by the distance it moves. The electric potential difference ΔV between points a and b is given by: Vb − Va = EL and work done Wab = q0 EL where: E is the strength of the electric field, L is the distance between points a and b, Wab is the work done. Figure 24.8: Motion of a charge in an equipotential surface from point a to b. Equipotential Surfaces: As we move around in space, the potential (electric potential energy per unit charge) changes. However, there are surfaces where the electric potential is the same everywhere. These are called equipotential surfaces, as shown in fig. 24.9. For instance, around a single point charge or between two oppositely charged plates, every point on an equipotential surface has the same electric potential.  Figure 24.9: Equipotential surfaces for three different situations. Creating Potential Differences: To create a potential difference between two points, we need a device like a battery. Batteries use chemical reactions (like those between carbon and zinc electrodes in sulfuric acid, as shown in fig. 24.10) to generate a potential difference. This difference in potential allows us to move charges and generate electric current. The potential difference ΔV produced by a battery can be calculated based on the chemistry involved and the physical setup of the battery. Figure 24.10: Carbon and zinc electrodes in sulfuric acid generates potential differences. Various Methods to Create Potential Difference: Potential differences can also be created by mechanical means, such as in the Van de Graaff generator, where charges are moved using belts and rollers against electric forces. Similarly, natural processes like ionization due to rising air currents can create potential differences, leading to phenomena like lightning. Problem: Consider two conducting balls with different charges. When connected by a wire, as shown in fig. 24.11, charge redistributes between them until they reach the same potential. The charge Q on each ball is proportional to its radius r and the potential V it reaches. Figure 24.11: Sphere of different radiuses having different surface charge densities are connected with conducting wire. Solution: as we know, surface charge density for a sphere is q = 4 R 2 for each sphere q1 q2 1 = and  2 = 4 R1 2 4 R 2 2  q1 =  1 4 R12 and q 2 =  2 4 R 2 2 using it in eq(4)  1 4 R12 R1 =  2 4 R 2 2 R2  1 R1 R2 2 =  2 R2 R12  R  1 = 2  2 R1 as given R2  R1 so we can say,  1   2 Thus, the electric field at the surface of the smaller sphere is greater than that at the surface of the larger sphere: Esmall sphere  Elarge sphere