#### Transcript AP Electric Potential

AP Physics: Electricity & Magnetism What is potential energy? The ability to do work The energy possessed by an object due to its position in a force field. Work is an energy transfer A constant force applied over a distance in the same direction. W Fd W sf si F ds U U i U f The work done by a conservative force equals the decrease in the potential energy of the particle. W W B A Fc ds U B A q 0E ds U U q 0 E ds B A This is called a line or path integral which is independent of the path taken from point A to B. B U A E ds V q0 The potential at any point in the field: This value is called Electric Potential and is independent of the value of q U V q0 Be careful, electric potential is not potential energy. Units? J/C The work per unit charge that an external agent must perform to move a test charge from A to B without a change in kinetic energy VP E ds P The electric field at an arbitrary point equals the work required per unit charge to bring a positive test charge from infinity to that point B U A E ds V q0 J N m V C C Joule Newton meter Volt Coulomb Coulomb The energy that an electron or proton gains or loses by moving through a potential difference of 1V 1.60210 J 1eV 19 What is the potential difference between two points if the displacement is parallel to the field lines? V VB VA E ds B A E cos 0ds Eds E ds B B B A A A V Ed VB VC Cross-section of equipotential surface A E B C The name given to any surface consisting of a continuous distribution of points having the same electric potential. No work is done in moving a test charge between any two points on an equipotential surface. Red lines are electric field lines Blue lines are equipotential surfaces Equipotential surfaces must be perpendicular to field lines. A negative charge gains electric potential energy when it moves in the direction of the electric field. Explain this. A proton loses potential energy when moving in the direction of the electric field, but picks up an equal amount of kinetic energy. Work done by a field is positive when energy is given to an object from the field (object is lowered in a gravitational field or moved in the direction of E filed lines). This occurs when Uf <Ui so, W=-ΔU V Ed d 0.0006m E 3 106 V /m V Ed V 3 106 V /m0.0006m V 1.8kV U V q V 120V v0 0 v? mp 1.67 1027 kg me 9.111031 kg e 1.60219 C 1 qV U KE mv2 2 2qV v m vp 1.5 10 m/s 5 ve 6.5 106 m/s qV U KE V 115V KE q V KE 7.371017 J q? q 6.4 1019 C VA 9V VB 5V q N ae (6.021023 )(1.6021019 C) q 96440.4C N a 6.02 1023 electrons W ? V VA VB V 9V (5V) V 14V W U qV W qV (96440.4C)(14V) W 1.35MJ What is the electric field due to a point charge (q)? Q in E dA 0 Field is radially outward (or inward), and the Gaussian surface is a sphere chosen to match symmetry of field. Q in EA 0 Q in q kq E 2 2 A0 4r 0 r kq E 2 r V VB VA r E ds rB A The following simplification can be made because the potential difference is not dependant upon the path taken through the field, only the endpoints. V r Edr rB A kq V r 2 dr A r rB 1 V kq r 2 dr A r rB 1rB V kq r rA rB 1 V kq r rA If we want to know the potential at one particular location near a point charge we chose the first point to be infinity. 1rB V kq r 1 1 V kq r kq V r kq V r Potential at a point due to many point charges. qi V k i ri This is a scalar value so it is much easier to calculate than E field. The potential energy of a pair of charged particles separated by a distance r12. this is also the work done to move q2 from infinity to point P where q1 is located. kq1q 2 U q 2V1 r12 V1 is the electric potential at point P due to charge q1. Notice that if the charges are like then positive work is done to bring the charges together (they have more energy together because they will repel). Energy delivered to the system requires positive work. Imagine q1 is at a fixed position. We want to bring q2 and q3 from infinity to a position near q1. q1q2 q1q3 q2q3 U k r13 r23 r12 Each term represents the the work required to bring each charge to a location near the other. It does not matter which order we bring them in from infinity. r12 q1 q2 r13 r23 q3 V E ds B A How can we use this to solve for E field? dV Eds If the E field only has one component… dV E x dx dV Ex dx dV Ex dx Remember that the potential is zero for displacements perpendicular to the field If your potential is a function of all three spatial coordinates (x, y, z) dV Ey dy dV Ez dz V 4x y y 2yz 2 2 To find the electric field, we must take the partial derivative of this potential… V Ex 4x2y y2 2yz x x Partial derivative: Treat other variables as constant while differentiating with respect to one variable. 2 2 4y2x 8xy E x 4x y 4y x x x V 4x y y 2yz 2 2 V 2 2 Ey 4x y y 2yz y y E y 4x2 2y 2z V Ez 2yz z z E z 2y E x,y,z 8xyˆi 4x2 2y 2zˆj 2ykˆ In vector notation this is often written as… ˆ ˆ ˆ E V i j k x y z is called the gradient operator If the potential is constant in some region, what is the electric field? If the electric field is zero in some region, what is the electric potential? If the potential is constant in some region, what is the electric field? If the electric field is zero in some region, what is the electric potential? v constant, E 0 E 0, v constant The difference in electric potential (voltage) measured when moving from point A to point B is equal to the work which would have to be done, against the electric field to per unit charge, move the charge from A to B. The electric potential at point P due to a continuous charge distribution can be calculated by dividing the charged body into segments of charge dq and summing the potential contributions over all segments. kq V r dq Vk r dq a P x dq Vk r r a x 2 2 All segments of charge are equidistant from point P V k a x 2 2 dq V kQ a x 2 2 dq a P The field is zero here, therefore the potential must be constant: kQ V a this potential is the work necessary to bring a test charge from infinity to the location in the center of the ring. This oddly shaped conductor with an excess positive charge is in equilibrium A B VB VA E ds 0 B A Eds 0 Along this surface path E is always perpendicular to the displacement ds. The surface of any charged conductor in equilibrium is an equipotential surface. Furthermore, since the electric field is zero inside the conductor, we conclude that the potential is constant everywhere inside the conductor and equal to its value at the surface. A B The electric field is large near points having small convex radii of curvature and reaches very high values at sharp points. B Because the potential on the surface of the cavity is an equipotential surface. A A cavity in a conductor with no charge in it…The electric field inside the cavity must be zero, regardless of the charge distribution on the outside surface of the conductor. Sensitive circuits and people can be protected if placed in a cavity inside of a conductor. Air can become a conductor as a result of the ionization or air molecules in regions of high electric fields. At STP this happens around 3x106 V/m +2Q . E . . . . A D B C -Q 2a -Q Three charges are arranged in an equilateral triangle, as shown. At which of these points (a,b,c bisect sides, d is equidistant from other points) is the electric potential smallest? +2Q . E . . . . A D B C -Q 2a -Q C) A small positive test charge will move towards an area of low potential. Ask yourself “Where would an small positive charge end up if released near these charges?” .P 10V 40V 70V The diagram shows a set of equipotential surfaces. At point P, what is the direction of the electric field? A) left B) right C) up the slide D) down the slide E) either left or right, which one cannot be determined .P 10V 40V 70V A) A small positive charge placed at P would move to a location of low potential (left). The force that moves it is caused by an electric field which will be in the same direction or opposite it. Equipotential surfaces are always perpendicular to electric field lines. a Q Q What is the electric potential at a point halfway between the two charges? A) kQ/a B) 2kQ/a D) Electric potential is a scalar C) zero D) 4kQ/a E) 8kQ/a V kQ kQ 2kQ 4kQ a a a a 2 2 2 A solid conducting sphere carries a charge +Q. Which of the following is true of the electric field E and the electric potential V inside the sphere? A) E=0 and V=0 B) E=0 and V≠0 C) E≠0 and V=0 D) E≠0 and B) The field in a conductor is always zero. E is the derivative of V, so to be zero, V must have been a constant. V≠0 E) It cannot be determined without knowing the radius of the sphere. dV E dr A negatively charged rod is brought near a metal object on an insulating stand, as shown. When charges stop moving, the left side of the object has an excess of positive charge, and the right side of the object, where the radius of curvature is less, has an excess of negative charge. Which of the following best describes the electric potential on the metal object? + ------- + + + - A) It is greatest on the + side B) It is greatest on the – side C) It is greatest at the center D) It is the same everywhere on the object E) It cannot be determined from the information given + ------- + + + - A) It is greatest on the + side B) It is greatest on the – side C) It is greatest at the center D) It is the same everywhere on the object E) It cannot be determined from the information given D) It is the same everywhere on the object. The surface of a conductor is an equipotential surface. The E field is 0 inside the conductor and V is constant. . L . Q L L Q L . Q L L . Q Four identical charged particles, each with charge Q, are fixed in place in the shape of an equilateral pyramid with sides of length L, as shown above. What is the potential energy of this arrangement of charges? . L In general: q1q 2 U 40r . Q Q L L L . Q L L . Q Each pair of charges is separated by a distance L. Each pair contributes: Q2 U 40L There are 6 pairings: 6Q 2 U 40L What is the amount of work required to assemble 4 identical point charges of magnitude Q at the corners of a square of side s? 1 1 1 1 1 1 U kQ s s s s 2s 2s 2 Q s Q 2s s Q Q 4 2 U kQ s 2s 2 kQ2 2 U 4 s 2 kQ 2 U 5.41 s A wire that has a uniform linear charge density λ is bent into the shape shown below. Find the electric potential at point O. 2R R . O 2R A positron is given an initial velocity of 6x106m/s. It travels into a uniform electric field, directed to the left. As the positron enters the field, its electric potential is zero. What is the electric potential at the point where the positron has a speed of 1x106m/s? + vi E