Coulomb s constant k = 9x10 9 N m 2 /C 2

Transcription

1 1 Part 2: Electric Potential 2.1: Potential (Voltage) & Potential Energy q 2 Potential Energy of Point Charges Symbol U mks units [Joules = J] q 1 r Two point charges share an electric potential energy of U = kq 1q 2 r Coulomb s constant k = 9x10 9 N m 2 /C 2 This potential energy can be used in the Law of Conservation of Energy where K is the kinetic energy of the charged particle. K i + U i = K f + U f An electron is moving at 10 6 m/s and is 1 nm from a heavy positive ion. How fast is the electron moving when it is 10 nm from the ion? The ion is singly ionized, i.e. one electron has been removed from the neutral atom. Assume that the ion is so heavy compared to the electron that the ion does not move. Ans. 7.4x10 5 m/s The following three charges are held fixed on an x-y grid. q 1 = +2 C at (0,0) q 2 = -4 C at (3 m,0) q 3 = +3 C at (0, 4 m) Find the potential energy of the group. Ans J

2 2 Voltage (Potential) Symbol V mks units [Volts = V] A charged object with charge q o produces a voltage at every point in space except at its position. This object shares a potential energy with another charge q. This energy is given by U = qv where V is the voltage of q o at the position of q. Voltage of Point Charges A charged object that can be approximated as a point charge produces a voltage V at point P where V = kq r Sign of voltage: If q is +, voltage is +. If q is -, voltage is - The following three charges are held fixed on an x-y grid. q r x P q 1 = +2 C at (0,0) q 2 = -4 C at (3 m,0) q 3 = +3 C at (0, 4 m) Note that this is the same charge distribution used in the previous example in the potential energy section. (a) Find the net voltage at the point (3 m, 4 m) to these point charges. (b) If a fourth charge of +2 C is placed at (3 m, 4 m), find its potential energy. Ans. (a) V (b) J

3 3 Equipotential Surfaces Every point on one equipotential surface is at the same voltage. If a charged particle moves from one equipotential surface to another surface at a different voltage, then the particle will change speed. The Law of Conservation of Energy can be rewritten as K i + qv i = K f + qv f Positive charges speed up when going from high to low voltage. They slow down when going from low to high voltage. Negative charges speed up when going from low to high voltage. They slow down when going from high to low voltage. Using the previous equation, you can write that the amount of kinetic energy gained or lost is K = q V = q V Note that in this equation, V is the absolute value of the voltage difference. Charged Parallel Plates (Uniform Electric Field) We already know that there will be a uniform electric field between a set of parallel plates with different amounts of charge. There is then a voltage difference V between the plates given by V = Ed where d is the distance between the plates. This voltage difference is usually is written as just plain V so the equation becomes V = Ed The electric field points from the more positive plate to the more negative plate. The more positive plate is at a higher voltage than the more negative plate. V is the difference between these two plate voltages. +Q + d E V _ -Q

4 4 Note from the above equation that the electric field can be expressed in units of [V/m]. This unit is equivalent to [N/C]. The motion of charged particles between the plates can be analyzed using the previous equation for the kinetic energy gained or lost by the particle as it travels across the plates. K = q V = q V The mks unit of energy is the Joule. The kinetic energies of typical charged particles is much less than a Joule so we introduce another energy unit called the electronvolt [ev] which is defined to be the kinetic energy gained by an electron if it is accelerated across 1 V of voltage difference. [1 ev = 1.6x10-19 J] Two parallel plates are separated by 2 cm. The top plate is at 90V and the bottom plate is at - 10V. You are instructed to accelerate an electron across the plates. Answer the following questions. (a) What is the size of the electric field between the plates? (b) Near which plate should you place the electron so it speeds up as it travels between the plates? (c) How much kinetic energy does the electron gain in traveling across the plates? Answer in units of Joules and electronvolts. (d) What is the final speed of the electron if it starts approximately at rest? Ans. (a) 5000 V/m (b) bottom plate (c) 1.6x10-17 J = 100 ev (d) 5.93x10 6 m/s

5 5 2.2 Capacitance & Capacitors A capacitor is a device that can store opposite charge (+Q and Q) on two different surfaces when a voltage difference (V) is applied between the surfaces. The capacitance (C) of the capacitor determines how much charge can be stored for a given voltage difference. Q = CV In this equation, Q is the absolute value of the charge on one of the surfaces and V is the voltage difference between the surfaces. mks units of capacitance [Farad = F = C/V] The capacitance of a capacitor depends on the shape of the capacitor and the material between the two surfaces. For a parallel-plate capacitor, A C = Κε oa d d where A is the area of one of the plates, d is the distance between the plates, and is the dielectric constant of the insulating material between the plates. The dielectric constant has no units and is a property of the material. A capacitor stores energy in the electric field between the plates with an amount equal to U = 1 2 CV2

6 6 Combinations of Capacitors Capacitors in parallel share the same voltage difference. The equivalent capacitance of two capacitors in parallel is C eq = C 1 + C Capacitors in series store the same amount of charge. The equivalent capacitance of two capacitors in series is given by 1 C eq = 1 C C 2 This equation can be solved for the equivalent capacitance to give C eq= C 1 C 2 C 1 + C 2 (a) Find the equivalent capacitance of the circuit. (b) Find for each capacitor the amount of stored charge, the voltage difference, and the amount of stored energy if a voltage difference of 12 V is applied between points A and B. A C 1 C 2 B C 3 C 1 = 3 F C 2 = 6 F C 3 = 4 F Ans. (a) 6 F (b) C 1 : 24 C, 8 V, 96 J C 2 : 24 C, 4 V, 48 J C 3 : 48 C, 12 V, 288 J

7 7 2.3 Current Symbol mks units I, i [Amperes = Amps = A = C/s] The current is the rate that charge flows from one point to another. I = Δq Δt By convention, current flows in the direction that positive charge would travel in a circuit. (In practice, it is the flow of electrons in the opposite direction that constitutes the current.) 2. 4 Resistance & Resistors When current flows through an object, there is always some resistance to this flow. A resistor is an object through which a current (I) flows when a voltage difference (V) is applied across the ends of the object. The resistance (R) of the resistor determines how much current flows for a given voltage difference. For most objects, I = V / R This equation is an expression of Ohm s Law which is usually written as V = IR mks units of resistance [Ohm = = V/A ] The resistance of a resistor depends on the shape of the resistor and the material from which it is made. R = ρ A A where A is the cross-sectional area of the object, is the object s length, and is the resistivity of the material. mks units of resistivity [ -m]

8 8 A resistor cannot store energy. Rather, it converts all of the electrical energy it receives into heat energy at a rate given by P = IV Recall that the rate that energy is delivered or used is the power P with units of Watts [ W = J/s]. Power = Energy / Time Thus, the amount of energy supplied or used is Energy = Power Time mks unit of energy [J = W s] common unit of electrical energy [kilowatt-hour = kw-hr] [1 kw-hr = 3.6x10 6 J] Combinations of Resistors Resistors in parallel share the same voltage difference. The equivalent resistance of two resistors in parallel is 1 = R eq R 1 R 2 This equation can be solved for the equivalent resistance to give R eq= R 1 R 2 R 1 + R Resistors in series have the same current flowing through each. The equivalent resistance of two resistors in series is given by R eq = R 1 + R 2

9 9 2.5 DC Circuits Resistor Circuits (a) Find the equivalent resistance of the circuit. (b) Find for each resistor the current, the voltage difference, and the dissipated power. R 1 40 V 8 R 2 R Ans. (a) 20 (b) R 1 : 2 A, 16 V, 32 W R 2 : 0.8 A, 24 V, 19.2 W R 3 : 1.2 A, 24 V, 28.8 W RC Circuits Charging Up a Capacitor: Charge on a plate increases in size, current decrease with time. Close switch at t = 0 R t τ q(t) = CV o (1 e ) I(t) = V o e t τ R Discharging a Capacitor: Both charge on a plate and current decrease with time. V o C t τ q(t) = Q o e I(t) = V o e t τ R C Close switch at t = 0 R In both cases, is the RC time constant: = RC [s] The capacitor is considered fully charge or discharged after about 5 time constants.

10 10 A 2- F capacitor is charged up with a 15-V battery through a resistance of (a) What is the RC time constant of the circuit? (b) What is the initial current that flows through the circuit? (c) What is the final charge stored by the capacitor? (d) Find the current and stored charge 2 milliseconds after the charging begins. (e) After charging up, the capacitor is discharged through the same resistance. How long does it take for the current to reach 1 ma? Ans. (a) 6 ms (b) 5 ma (c) 30 C (d) 3.58 ma and 8.52 C (e) 9.7 ms