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## Lecture 4 Work, Electric Potential and Potential Energy Ch. 25

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**Lecture 4 Work, Electric Potential and Potential Energy Ch.**25 • Topics • Work, electric potential energy and electric potential • Calculation of potential from field • Potential from a point charge • Potential due to a group of point charges, electric dipole • Potential due to continuous charged distributions • Calculating the electric field from a potential • Electric potential energy from a system of point charges • Equipotential Surface • Potential of a charged isolated conductor • Demos • Teflon and silk • Charge Tester, non-spherical conductor, compare charge density at different radii. • Elmo • Potential in the center of four charges • Potential of a electric dipole • Polling**Work, Potential Energy and Electric potential**• The electric force is mathematically the same as gravity so it too must be a conservative force. We want to show that the work done is independent of the path and only depends on the endpoints. Then the force is said to be a conservative force. • First start with work Work done by the electric force = • Then we will find it useful to define a potential energy as is the case for gravity. • And the electric potential**b**P2 P3 c a E P1 Lets start with a uniform electric field and find the work done for a positive test charge.**b**P2 P3 F c a E P1 Find work done along path W12 for a positive test charge**F**b P2 P3 c a E P1 Find Work along path W23**W12 + W23 = 0 + qEb =qEb**Compare this work done along path W13**Work done along path W13**b P2 P3 c a F P1**Conclusion**• Work done along path W12 + W23 = W13. • Work is independent of the particular path. • Although we proved it for a uniform field, it is true for any field that is a only a function of r and is along r. • It only depends on the end points i and f. • This means we can define a function at every point in space and when we take the difference of that function between any two points, it is equal to the negative of the work done.**b**P2 P3 c a F P1 When we go from P1 to P3 we evaluate the Work function at P3 and subtract the value at P1 and then the a difference equals the negative of the work done in going form P1 to P3. This function is called the potential energy function**Example of finding the Potential Energy Function U in a**Uniform Field E E d b a What is the electric potential difference for a unit positive charge moving in an uniform electric field from a to b? If we set the origin at xb = 0, and measure from b to a, then Ub=0 and the potential energy function is U=qEd This is analogy with gravitation where we U =mgh.**Now define the Electric Potential Difference V which does**not depend on charge. • The potential difference is the negative of the work done per unit charge by an electric field on a positive unit charge when it moves from one point to another.**Find the potential difference V for a**uniform electric field For a battery of potential difference of 9 volts you would say that the positive terminal is 9 volts above the negative terminal.**Note relationship between potential and electric field**• V is a scalar not a vector. Simplifies solving problems. • We are free to choose V to be 0 at any location. Normally V is chosen to be 0 at the negative terminal of a battery or 0 at infinity for a point charge.**Example for a battery in a circuit**• In a 9 volt battery, typically used in IC circuits, the positive terminal has a potential 9 v higher than the negative terminal. If one micro-Coulomb of positive charge flows through an external circuit from the positive to negative terminal, how much has its potential energy been changed? q**Generalize concept of electric potential energy and**potential difference for any electric field**= - Work done by the electric force =**y x (independent of path, ds) Therefore, electric force is a conservative force.**Find the electric potential when moving from one point to**another in a field due to a point charge?**Potential of a point charge at a distance R**Replace R with r eqn 25-26**Electric potential for a positive point charge**• V is a scalar • V is positive for positive charges, negative for negative charges. • r is always positive. • For many point charges, the potential at a point in space is the simple algebraic sum (Not a vector sum)**Electric potential due to a positive point charge**Hydrogen atom. • What is the electric potential at a distance of 0.529 A from the proton? 1A= 10-10 m r = 0.529 A • What is the electric potential energy of the electron at that point? U = qV= (-1.6 x 10-19 C) (27.2 V) = - 43.52 x 10-19 J or - 27.2 eV where eV stands for electron volt. Note that the total energy E of the electron in the ground state of hydrogen is - 13.6 eV Also U= 2E = -27.2 eV. This agrees with above formula.**What is the electric potential due to several point charges?**For many point charges, the potential at a point in space is the simple algebraic sum (Not a vector sum) y q2 q1 r2 r1 q3 r3 x**Two point charges that are opposite and equal**What is the potential due to a dipole?**Potential for a Continuous Distribution of Charge**Point charge For an element of charge Integrate V**Chaper 24 Problem 22. With V = 0 at infinity, what is the**electric potential at P, the center of curvature of the uniformly charged nonconducting rod?**Chapter 24 Problem 26. What is the magnitude of the net**electric potential at the center? 1. A thin rod of charge -3.0 µC that forms a full circle of radius 6.0 cm2. A second thin rod of charge 2.0 µC that forms a circular arc of radius 4.0 cm, subtending an angle of 90° about the center of the full circle3. An electric dipole with a dipole moment that is perpendicular to a radial line andthat has magnitude 1.28 multiplied by 10-21 C·m**Figure 24-44 shows a thin plastic rod of length L and**uniform positive charge Q lying on an x axis. With V = 0 at infinity, find the electric potential at point P1 on the axis, at distance d from one end of the rod.**Potential due to a ring of charge**• Direct integration. Since V is a scalar, it is easier to evaluate V than E. • Find V on the axis of a ring of total charge Q. Use the formula for a point charge, but replace q with elemental charge dq and integrate. Point charge For an element of charge r is a constant as we integrate. V This is simpler than finding E because V is not a vector.**Potential due to a line charge**We know that for an element of charge dq the potential is For the line charge let the charge density be l. Then dq=ldx Now, we can find the total potential V produced by the rod at point P by integrating along the length of the rod from x=0 to x=L**A new method to find E if the potential is known.If we know**V, how do we find E? So the x component of E is the derivative of V with respect to x, etc. • If V = a constant, then Ex = 0. The lines or surfaces on which V remains constant are called equipotential lines or surfaces. • See example on next slide**Now find the electric field at point P1 on the axis, at**distance d from one end of the rod. Find the x and y components Ex and Ey.**Equipotential Surfaces**• Three examples • What is the equipotential surface and equipotential volume for an arbitrary shaped charged conductor? • See physlet 9.3.2 Which equipotential surfaces fit the field lines?**x**Blue lines are the electric field lines Orange dotted lines represent the equipotential surfaces a) Uniform E field V = constant in y and z directions c) Electric Dipole (ellipsoidal concentric shells) • Point charge (concentric shells)**Electric Potential Energy U of a system of charges**How much work is required to set up the arrangement of Figure 24-46 if q = 3.20 pC, a = 54.0 cm, and the particles are initially infinitely far apart and at rest? q2 W = U q1 q3 q4**Dielectric Breakdown: Application of Gauss’s Law**If the electric field in a gas exceeds a certain value, the gas breaks down and you get a spark or lightning bolt if the gas is air. In dry air at STP, you get a spark when V = constant on surface of conductor Radius r2 r1 1 2**This explains why:**• Sharp points on conductors have the highest electric fields and cause corona discharge or sparks. • Pick up the most charge with charge tester from the pointy regions of the non-spherical conductor. • Use non-spherical metal conductor charged with teflon rod. Show variation of charge across surface with charge tester. Radius R V = constant on surface of conductor 1 Cloud 2 + + + + Van de Graaff - - - -**How does a conductor shield the interior from an exterior**electric field? • Start out with a uniform electric field with no excess charge on conductor. Electrons on surface of conductor adjust so that: 1. E=0 inside conductor 2. Electric field lines are perpendicular to the surface. Suppose they weren’t? 3. Does E = just outside the conductor 4. Is s uniform over the surface? 5. Is the surface an equipotential? 6. If the surface had an excess charge, how would your answers change?**What is the electric potential of a uniformly charged**circular disk? We can treat the disk as a set of ring charges. The ring of radius R’ and thickness dR’ has an area of 2pR’dR’ and it’s charge is dq = sdA = s(2pR’)dR’ where s=Q/(pR2), the surface charge density. The potential dV at point P due to the charge dq on this ring given by q Integrating R’ from R’=0 to R’=R**Chapter 24 Problem 19**The ammonia molecule NH3 has a permanent electric dipole moment equal to 1.31 D, where 1 D = 1 debye unit = 3.34 multiplied by 10-30 C·m. Calculate the electric potential due to an ammonia molecule at a point 44.0 nm away along the axis of the dipole. (Set V = 0 at infinity.)**Chapter 24 Problem 55**Two metal spheres, each of radius 1.0 cm, have a center-to-center separation of 2.2 m. Sphere 1 has charge +2.0 multiplied by 10-8 C. Sphere 2 has charge of -3.8 multiplied by 10-8 C. Assume that the separation is large enough for us to assume that the charge on each sphere is uniformly distributed (the spheres do not affect each other). Take V = 0 at infinity. (a) Calculate the potential at the point halfway between the centers. (b) Calculate the potential on the surface of sphere 1. (c) Calculate the potential on the surface of sphere 2.**Chapter 24 Problem 57**A metal sphere of radius 11 cm has a net charge of 2.0 multiplied by 10-8 C. (a) What is the electric field at the sphere's surface? (b) If V = 0 at infinity, what is the electric potential at the sphere's surface? (c) At what distance from the sphere's surface has the electric potential decreased by 500 V?