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Physics 8.02 F= Quiz One Equations 1 q1q2 4 !" o r 2 C= ! E= q q ! rˆ = r 2 4!" o r 4!" o r 3 ! r rˆ = points from source q to observer r Q !V Fall 2006 2 Q2 U = 1 C !V = 2 2C C parallel = C1 + C2 Cseries = C1C2 C1 + C2 Circumferences, Areas, Volumes: ! 1 dq E= rˆ # 4!" 0 V r 2 1) The area of a circle of radius r is π r2 ! ! Q ! E " dA = inside $o closed surface 2) The surface area of a sphere of radius r is 4π r2 Its circumference is 2π r " ## ! dA points from inside to outside Its volume is ! ! !V = V f " Vi = " $ E # dr f " ! ! E ! dr = 0 i 3) The area of the sides of a cylinder of radius r and height h is 2π r h closed path Vpoint charge = Its volume is π r2 h q 4!" o r N Vmany point charges = $ i =1 U= Definition of trig functions qi ! " 4!" o r # ri qi q j ! ! all pairs 4#$ o ri " r j ! "1 2% U=! ((( $# 2 ! o E '& dVvol !V !V !V Ex = , Ey = , Ez = !x !y !z !V Er = " for spherical symmetry !r sin is opposite/hypotenuse; cos is adjacent/hypotenuse; tangent is opposite over adjacent; Properties of 30, 45, and 60 degrees (π/6, π/4, and π/3 radians): sin(π/6) = cos(π/3) = 1/2, sin(π/3) = cos(π/6) = 3 / 2 ; sin(π/4) = cos(π/4) = 1 / 2 ; Integrals that may be useful b " dr = b ! a a b 1 ! r dr = ln(b / a ) a b 1 (r a 1 4 3 !r 3 2 &1 1# dr = $ ' ! %a b" MIT PHYSICS DEPARTMENT Physics 8.02 ` Quiz One Solutions Fall 2006 Part I Concept Questions: (50 points) (a) (10 points) A point charge is placed outside a hollow conducting sphere that was initially uncharged. Which of the following statements are true or false? 1. The electric field inside the sphere is zero. 2. There is a (possibly non-constant) surface charge density on the inner surface of the sphere. 3. There is a (possibly non-constant) surface charge density on the outer surface of the sphere. Answer: 1 and 3 are true, 2 is false. The point charge outside the sphere will induce a non-uniform surface charge density on the outer surface of the sphere. The superposition of this electric field and the electric field of the charge outside will give a zero electric field everywhere inside the hollow spherical conductor. Denote by a the inner radius of the hollow conducting sphere and b the outer radius. By Gauss’s Law, the electric flux is zero through a Gaussian spherical surface that has a radius r such that a < r < b . Therefore the total surface charge density on the inner surface of the sphere is zero. Since the inner surface is screened from the charge outside, the only possibility is that there is zero charge density everywhere on the inner surface since there is no field that can separate the charges. An energy argument also gives the same result. The lowest energy configuration is for all the charge to reside on the outer surface and the electric field everywhere inside the outer surface zero. 2 MIT PHYSICS DEPARTMENT ` (b) (10 points) Circle your answer. If you place a negatively charged particle in an electric field, the charge will move 1. from higher to lower electric potential and from lower to higher potential energy. 2. from higher to lower electric potential and from higher to lower potential energy. 3. from lower to higher electric potential and from lower to higher potential energy. 4. from lower to higher electric potential and from higher to lower potential energy. Answer: 4. From lower to higher potential and higher to lower potential energy. Objects always move to reduce their potential energy, !U " U f # U i < 0 . Negative charges do this by moving towards a higher potential. The electric potential difference is defined as the potential energy difference per charge. !V = !U / q Since !U < 0 (always) and q < 0 (negative charge), !V " V f # Vi > 0 . So V f > Vi . 3 MIT PHYSICS DEPARTMENT ` (c) (10 points) Circle your answer. Two opposite charges are placed on a line as shown in the figure below. The charge on the right is three times the magnitude of the charge on the left. Besides infinite, where else can electric field possibly be zero? 1. between the two charges. 2. to the right of the charge on the right. 3. to the left of the charge on the left. 4. the electric field is nowhere zero. ! Answer: 3. The electric field from the positive charge E+3q and the electric field ! from the negative charge E! q are shown for different regions with the arrow size indicative of the magnitude in the figure below. The only place the fields can cancel is to the left of the negative charge. 4 ` MIT PHYSICS DEPARTMENT (d) (10 points) Circle your answer. Consider a parallel plate capacitor with circular plates of area A separated by a distance d. Suppose we introduce a conducting circular plate of area A parallel to the first two plates halfway between the plates. What happens to the capacitance: 1) It doubles. 2) It halves. 3) It is unchanged. Answer: 3. Unchanged. Two identical capacitors in series add inversely, 1 1 1 2 = + = . Ceq C C C So the equivalent capacitance has halved Ceq = C / 2 . A parallel plate capacitor with circular plates of area A separated by a distance d with charge per area ! has capacitance C= Q "A = !V E d The electric field of oppositely charged parallel plates is approximately given by Gauss’s Law E = ! / " 0 . So the capacitance is C0 = # A Q "A = = 0 . !V (" / # 0 ) d d If the distance between the plates is halved then the capacitance doubles C= 2! 0 A . d So the equivalent capacitance Ceq = C / 2 = !0 A = C0 . d From an energy perspective, the added plate has negligible thickness, so the electric field still occupies the same space. Therefore the energy stored is unchanged hence the capacitance is unchanged by U = Q 2 / 2C . 5 MIT PHYSICS DEPARTMENT ` (e) (10 points) Circle your answer. Two charges lie on the x-axis. A representation of the equipotentials of the electric potential of these two charges is shown above. The “streaks” in this representation are parallel to the equipotential curves. Which one of the following statements is true. 1. The two charges have opposite signs and the charge on the left is smaller in magnitude than the charge on the right. 2. The two charges have opposite signs and the charge on the left is larger in magnitude than the charge on the right. 3. The two charges have the same sign and the charge on the left is smaller in magnitude than the charge on the right. 4. The two charges have the same sign and the charge on the left is larger in magnitude than the charge on the right. Answer: 2. The equipotential line show a zero point to the right of the charge on the right. Therefore the two charges must be of opposite signs The charge on the left is further away from the zero point so the magnitude of the charge must be larger than the charge on the right in order to cancel the potential from the charge on the right. 6 MIT PHYSICS DEPARTMENT ` Part II Analytic problem: (50 points) An infinite slab of charge carries a charge per unit volume ρ, as shown in the sketch on the left. The plot on the right shows the electric potential V (x) in volts due to this slab of charge as a function of horizontal distance x from the center of the slab. The “slab” is 2 meters wide in the x-direction, and its boundaries are located at x = -1 meter and x = +1 meter, as indicated. The slab is infinite in the y direction and in the z direction (out of the page). (a) The potential V (x) is a linear function of x in the region x < -1 meter. What is the xcomponent Ex of the electric field in the region x < -1 meter? Be careful to indicate the correct sign and units of your answer. Ex = " !V "15 volts =" = 15 volts / meter !x 1 meter (b) The potential V (x) is a linear function of x in the region x > 1 meter. What is the xcomponent Ex of the electric field in the region x > 1 meter. Be careful to indicate the correct sign and units of your answer. Ex = " !V 15 volts =" = "15 volts / meter !x 1 meter 7 ` MIT PHYSICS DEPARTMENT (c) In the region –1 meter < x < 1 meter, the potential V (x) is a quadratic function of x given by the equation " 15 % " 25 % V (x) = $ V ! m -2 ' x 2 ( $ V . # 2 & # 2 '& What is the x-component of the electric field Ex in this region? Be careful to indicate the correct sign and units of your answer. Ex = ! "V = !15x volts / meter "x (d) Use Gauss’s Law and your answers above to find the charge density ρ of the slab. You must indicate the Gaussian surface you use on the figure below. You can use the symbol εo in your answer. Be careful to indicate the correct sign and units of your answer. Take a cylindrical gaussian pill box spanning the slab with one face in the empty space to the left and one face in the empty space to the right. We have the electric fields on the faces of this pillbox from above, and we find that ! ! Qinside E " "" ! dA = # o = $15A $ 15A = A% 2 / # o closed surface where A is the area of the face of the pillbox. So ! = "# o 15coulombs / cubic meter 8 ` MIT PHYSICS DEPARTMENT (e) How much work do you do to move 1 micro-Coulomb from: (i) x = 1 m to x = 2 m . (ii) x = !1 m to x = 1 m (iii) x = !2 m to x = !1 m (iv) x = 1 m to x = infinity (i) , (iii), and (iv) . Zero. For each of these cases, the potential is constant in each region so the potential difference between any two points in each region is zero, so the work done is zero. (ii) Zero. Since the electric field changes direction in side this region, the work done between any two points that are equidistant from the center is zero. This makes sense because half the time you are doing positive work to move the charge, and half the time you are doing negative work, and the net work turns out to be zero. From the definition of work done by the field in moving between two points x =1 m W= # ! ! F ! dr = x = "1 m x =1 m # ! ! qE ! dr x = "1 m x =1 m W =q " Ex dx x = !1 m x =1 m x2 W = (1 ! 10 C) $ "(15 V # m )x dx = "(1 ! 10 C)(15 V # m ) 2 x = "1 m "6 -1 "6 x =1 m -1 x = "1 m W = !(1 " 10 !6 C)(15 V # m -1 )(1 / 2 ! 1 / 2) = 0 9