PHY 2049 Spring 2013 Exam 1 solutions Exam 1 Solutions Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1 What is the total enclosed charge in the shown cube if the electric field is given by E = −3î + y 2 ĵ (or E = −3î + 2y 2 ĵ ) and the cube has a side length of 2? (a) 16 ε0 (b) 0 (c) 4 ε0 (d) 32 ε0 (e) ε0 Because of the directions of the electric field, the flux could only be non-‐zero for the top-‐bottom and left-‐right faces. However, because the field also depends on y, the only non-‐zero contribution is for the top face. The left and right faces have a constant field and so no net flux. qenc = ε 0 Φ = ε 0 ΦT = ε 0 ET ⋅ A T = ε 0 (2)2 (2)2 = 16ε 0 or = ε 0 2(2)2 (2)2 = 32ε 0 Recall that flux is positive when the field points out from a closed surface. Problem 2 What is the x component of the electric field at the center of the square array of charged particles with q1 = +2q , q2 = +2q (or +3q ) , q3 = +2q , and q4 = +q ? The side length a = 10 cm and q = 2 x 10-‐9 C (a) -‐2550 N/C (b) 3600 N/C (c) -‐5100 N/C (d) 1270 N/C (e) 0 Note that the electric field contributions from q1 and q3 cancel since the charges are equal. The charges q2 and q4 are unequal, so the net imbalance is +q (or +2q) in the location of q2. PHY 2049 Spring 2013 Exam 1 solutions Ex = −K (a q 2 /2 ) 2 2 = −2550 N/C 2 or Ex = −K (a 2q 2 /2 ) 2 2 = −5100 N/C 2 Don’t forget the field points away from + charge, and you have to project onto the x axis. Problem 3 Two identical conducting spheres are separated by a fixed distance of d=30cm, which is much larger than the sphere radius. The sphere charges are initially q1 = +1.0nC and q2 = -‐3.0nC (or -‐4.0nC or -‐5.0nC). They are then connected by a thin conducting wire and then disconnected. What is the magnitude of the electrostatic force in N between the spheres? (1) 10-‐7 (2) 2x10-‐7 (3) 3x10-‐7 (4) 4x10-‐7 (5) 5x10-‐7 When the wire is connected, the total net charge is shared and split equally between the two identical spheres. Therefore after disconnecting, each sphere will have a charge of (+1 – 3)/2 = -‐1 nC (or -‐1.5 nC or -‐2.0 nC). The electric force is then: (10 ) = 10−7 N or 2.25x10−7 N or 4x10−7 N qq F = K 1 2 2 = ( 9 × 10 9 ) d ( 0.3)2 −9 2 Problem 4 A particle with charge q1=-‐0.4e is at x=0 and another particle with charge q2=+3e is at x=0.63m. At what location along the x axis and with what charge could a third particle be placed in order to bring the entire system into electrostatic equilibrium (all particles stationary)? (1) x=-‐0.37m, q=+e (2) x=+1.0m, q=-‐e (3) x=+1.26m, q=-‐0.4e (4) x=-‐0.63m, q=+3e (5) x=-‐0.21m, q=+e The simplest way to solve this is to just try each choice for the third particle and see if all forces on each particle vanish. For example, let’s find the net force on particle 1. ⎛ qq qq ⎞ F1,net = F12 + F13 = K ⎜ 1 22 ± 1 2 3 ⎟ , where the ± sign depends on the direction x ⎠ ⎝ 0.63 of the force from the third particle. PHY 2049 Spring 2013 Exam 1 solutions The force from the presence of particle 2 is attractive, so in the +x direction. If another positively charged particle is placed with x < 0, the force also would be attractive but in the –x direction, potentially cancelling the force from the second particle by choosing the “-‐“ sign for “±”. We see that by substituting in x=-‐0.37m, q3=+e yields a net force that (approximately) vanishes. (and it does so also for the net force on particles 2 and 3). Some of the other choices might lead to the particle at x=0 to be stationary, but the outer charges would not be. Problem 5 An electron is positioned 0.5cm above a large, flat non-‐conducting sheet of charge of surface charge density σ = +2.0 μC/m2. When released from rest, how long does it take the electron to reach the sheet? (me = 9.11x 10-‐31kg, e=1.6x 10-‐19C). (1) 7 x 10-‐10s (2) 2 x 1016s (3) 3 x 10-‐5s (4) 105s (5) 5 x10-‐19s σ The electric field from a large sheet is E = 2ε 0 The electron is negatively charged and accelerates toward it with acceleration: F eE eσ a= = = me me 2 ε 0 me The distance it falls is 0.5cm, so we can solve for the time: 1 d = 0.005m = at 2 2 2d2ε 0 me −10 ⇒t = = 7 × 10 s eσ In case you worry about neglecting gravity, the acceleration itself from the electric field is 2 × 1016 m/s2, so yes it can be neglected. PHY 2049 Spring 2013 Exam 1 solutions Problem 6 A glass rod forms a quarter circle of radius R=2.5cm with a charge of q=+10-‐11C distributed uniformly along it. What is the x component of the electric field at the center of the arc (the origin of the axes shown)? (1) -‐92 N/C (2) -‐144 N/C (3) 520 N/C (4) -‐23 N/C (5) 0 The direction of the electric field from the quarter circle would be down and to the left at 45 degrees by symmetry. So the x component would be negative. The magnitude is given by: q λ= = 2.55 × 10 −10 C/m π R 2 dq = λ ds = λ Rdθ π /2 dq π /2 λ Rdθ Ex = − ∫ k 2 cosθ = − ∫ k cosθ 0 0 R R2 λ λ = −k sin θ |π0 /2 = −k R R = −92 N/C PHY 2049 Spring 2013 Exam 1 solutions Problem 7 A rod of charge per unit length λ1 is surrounded by a thin, concentric cylinder of charge per unit length λ2 and radius R (see figure). What is the magnitude of the electric field at a radius r<R (or r>R)? (1) λ1 / (2 π ε0 r) (2) (λ1 + λ2) / (2 π ε0 r) (3) (λ1 -‐ λ2) / (2 π ε0 r) (4) λ2 / (2 π ε0 r) (5) 0 Use Gauss’ Law to solve for the electric field based on the charge enclosed: Φ= ∫ E ⋅ d A = E ( r ) ∫ dA = E ( r ) 2π rL S S since E is constant at a fixed radius. Then set this equal to the charge enclosed over ε0. ⎧ λ1 L ⎪ ,r<R ε0 qenc ⎪ Φ = E ( r ) 2π rL = =⎨ ε 0 ⎪ ( λ1 + λ2 ) L ,r>R ⎪ ε0 ⎩ ⎧ λ1 ⎪ ,r<R ⎪ 2πε 0 r ⇒ E (r ) = ⎨ ⎪ ( λ1 + λ2 ) , r > R ⎪ 2πε r 0 ⎩ PHY 2049 Spring 2013 Exam 1 solutions Problem 8 What is the equivalent capacitance in µF of the circuit shown if each capacitor has a capacitance of 1μF? (1) 6/11 (2) 11/6 (3) 6 (4) 5/6 (5) 1/6 Capacitors C2, C3, and C4 are in parallel with each other. Therefore their capacitances add, giving 3 μF. Similarly capacitors C5 and C6 are in parallel with each other, giving capacitance 2 μF. This then gives us 3 capacitors in series with capacitances 1 μF, 2 μF, and 3 μF. So the total capacitance is: −1 −1 6 ⎛ 1 1 1⎞ ⎛ 6 + 3+ 2⎞ Ceq = ⎜ + + ⎟ = ⎜ = µF ⎟ ⎝ 1 2 3⎠ ⎝ ⎠ 6 11 Problem 9 In the shown figure, a potential difference of V = 12V is applied across the arrangement of capacitors with capacitances of C1 = C2 = 4μF, and C3 = 1μF. What is the charge q3 on one of the plates of capacitor C3? (1) 12μC (2) 24μC (3) 36μC (4) 6μC (5) 48μC The voltage applied across capacitor C3 is V. Therefore the charge q3 on its plates is q3 = CV = (12V) (1µF ) = 12 µC No need to compute the equivalent capacitance of the circuit. PHY 2049 Spring 2013 Exam 1 solutions Problem 10 The figure gives the magnitude of the electric field inside and outside a solid sphere with a positive charge distributed uniformly throughout its volume. The scale of the vertical axis is set by Es = 5.0 x 107 N/C. What is the total charge of the sphere? (1) 2.2μC (2) 110μC (3) 0.5μC (4) 55μC (5) 0 The diagram shows the electric field magnitude both inside and outside of a solid sphere with a uniform charge density. The change in shape at r=2cm indicates the outer radius of the sphere. Outside a sphere, the electric field has the magnitude of that of a point charge at the center of the sphere: Q E = K 2 r So we can solve for the charge Q by using the field magnitude at r=2cm: 5 × 10 7 N/C ) ( E 2 Q= r = ( 0.02m )2 = 2.2 µC 9 2 2 K ( 9 × 10 Nm / C ) Problem 11 An air-‐filled parallel-‐plate capacitor has a capacitance of 1μF and has a potential difference of 100V between the plates. If the gap between the plates is inserted with a dielectric material with dielectric constant κ = 2.5, how much additional energy is required to do this if the potential difference is maintained at 100V? (1) 7.5 x 10-‐3 J (2) 1.25 x 10-‐2 J (3) 5.0 x 10-‐3 J (4) 2.0 x 10-‐3 J (5) 0 Let’s calculate the energy for each configuration: 1 1 U1 = C1V 2 U 2 = C2V 2 2 2 Now the capacitance of the parallel plate capacitor with a dielectric is: C2 = κ C1 . So 1 ΔU = U 2 − U1 = (κ − 1) C1V 2 2 1 2 −6 −3 = (1.5 ) 1× 10 F (100 V) = 7.5 × 10 J 2 ( ) PHY 2049 Spring 2013 Exam 1 solutions Problem 12 Consider the electric dipole shown in the figure, where the distance r to point P is much larger than the dipole separation distance d. If the distance r is doubled, what is the ratio of the magnitude of the electric field at the new location to that at the original position? (1) 1/8 (2) 8 (3) ¼ (4) 4 (5) ½ The total electric field is the sum of the contributions from each charge. For large r this was shown to fall-‐off as r-‐3: qd ETOT ≈ K 3 r Therefore, when the distance is doubled, the field magnitude is 1/8 of the initial value. Problem 13 Consider a non-‐conducting horizontal plane with a charge density of σ= 3 nC/m2. A small ball of charge q = 2 nC floats above the plane in the Earth's gravitational field. What is the ball's mass? (A) 3.45 x 10-‐8 kg. (B) 6.92 x 10-‐8 kg. (C) 5.51 x 10-‐9 kg (D) 8.69 x 10-‐7 kg (E) 6.92 x 10-‐4 kg The electric field above the plate points upward with magnitude E =σ/2ε0, hence the electrostatic force on the ball is Fe = q σ/2ε0 upward. The only other force is gravity, which points down and has magnitude Fg = mg. For the ball not to move we must have Fe = Fg, which implies m = q σ/2 g ε0. PHY 2049 Spring 2013 Exam 1 solutions Problem 14 A cylinder of radius 1 cm contains a negative charge density whose magnitude grows linearly with radius from zero with slope 5 pC/m4. What is the direction and magnitude of the electric field 0.5 cm from the central axis? (A) inward, 4.71 x 10-‐6 N/C (B) outward, 4.71 x 10-‐6 N/C (C) none, zero (D) inward, 7.06 x 10-‐6 N/C (E) outward, 7.06 x 10-‐6 N/C Use Gauss's law: In step one we use symmetry to infer that the electric field is radial, with magnitude E(r), where r is the distance from the central axis. In step 2 we choose the surface to be a cylinder of radius r and length L. This makes the flux Φ = 2π L r E(r). In step 3 we compute the charge enclosed in the surface. To grow linearly (from zero) with r the negative charge density must have the form ρ = -‐ρ' r, where ρ' = 5 pC/m4. Hence the charge enclosed is: r 2π L Q = L ∫ ( − ρ ' x )2π xdx = − ρ ′r 3 0 3 Now use Gauss's law to set Φ = Q/ε0, implying ρ′2 E (r ) = − 3ε 0 Problem 15 The electric field vector is E ( x, y ) = îa + ĵb , where a and b are constants. What is the electric potential V(x,y) for this field? (A) -‐a x -‐ b y (B) a x + b y (C) a x (D) –ab (E) -‐a b x y Recall that the electric field is minus the gradient of the potential. Start with the x component: ∂V Ex = a = − ∂x This implies V(x,y) = -‐a x + C(y), where C(y) is an arbitrary function of y. PHY 2049 Spring 2013 Exam 1 solutions Now use the y component: ∂V Ey = b = − = −C ' ( y ) ∂y This implies C(y) = -‐b y + const. Hence we have V(x,y) = -‐a x -‐ b y + const. The constant is not fixed but only the first answer has the correct dependence upon x and y. Problem 16 The electric potential is V(x,y) = A sin(k x) cos(k y), where A and k are constants. What is the y component of the electric field? (A) A k sin(k x) sin(k y) (B) -‐A k sin(k x) sin(k y) (C) A/k sin(k x) cos(k y) (D) A k sin(k y) (E) A cos(k y) Recall again that the electric field is minus the gradient of the potential. Hence ∂V Ey = b = − = −Asin ( kx ) ( −k ) sin ( ky ) = kA in ( kx ) in ( ky ) ∂y Problem 17 Four identical charges q are arranged in a square of side length d. What is the electrostatic energy of this system in units of q2/ε0d? (A) 0.43 (B) 0.48 (C) 0.32 (D) 0.40 (E) 0.22 Label the charges 1, 2, 3, and 4, proceeding counter-‐clockwise around the qq square. Each pair of charges contributes to the potential energy as i j 4πε 0 rij where rij is the distance from qi to qj. The pairs 12, 23, 34, and 41 are all a distance d apart; pairs 13 and 24 are a distance 2d apart. Hence the total potential energy is q2 ⎛ 1 1 ⎞ q2 ⎛ 4 + 2 ⎞ q2 U= + = 0.43 ⎜ 1+ 1+ 1+ 1+ ⎟= 4πε 0 d ⎝ ε 0d 2 2 ⎠ ε 0 d ⎜⎝ 4π ⎟⎠ PHY 2049 Spring 2013 Exam 1 solutions Problem 18 A conducting cylinder of radius 2 cm contains 4 pC/m of Charge along its length. An electron is released from rest 5 cm from the central axis. What is its kinetic energy in electron volts when it hits the cylinder? (A) 0.0659 (B) 0.0719 (C) 0.132 (D) 0.414 (E) It never hits By Gauss's law the electric field points outward with magnitude E = λ / (2πε0r). Hence the electric potential difference from r=a to r=b is Vab = -‐λ/(2πε0) ln(b/a). The electron starts at a = 5 cm and moves to b = 2 cm, so it loses a potential energy of (e λ) / (2 π ε0) ln(5/2), which becomes its kinetic energy. It is also useful to recall that one volt equals 1 J/C = 1 N-‐m/C, i.e. 1 electron volt = e(1 V) = 1.6x10-‐19 J. Problem 19 A capacitor is constructed by stacking three large conducting plates, each with surface area A, on top of each other. The middle plate is suspended a distance d above the lower plate. The top plate is suspended a distance 2d above the middle plate. If the top plate is maintained at potential V, and the bottom plate is at zero potential, how much charge is there on the bottom surface of the top plate? Express your answer as a number times ε0 A V/d. (A) 1/3 (B) ½ (C) 1 (D) 2 (E) 3 Recall that a parallel plate capacitor of area A and height h has capacitance C = ε0 A/h. The top stack has capacitance C1 = ε0 A/2d and the bottom stack has C2 = ε0 A/d. They are in series so the total capacitance is Ceq = (C1 C2)/(C1 + C2) = (1/3 ) ε0 A/d . Hence the charge must be Q = Ceq V = (1/3) ε0 A V / d. PHY 2049 Spring 2013 Exam 1 solutions Problem 20 Consider the capacitor formed by a conducting sphere of radius R, with the other surface at infinity. What is the capacitance of this system? (A) 4 π ε0 R (B) 4π ε0 (C) 4π ε0/R (D) 2π ε0 R (E) 2π ε0/R Recall how to compute capacitance of two conducting surfaces: Place charge ±Q on the two surfaces, compute the potential difference between them, and then take C = Q/V. If we place a charge Q on the sphere then its potential difference with respect to the surface at infinity is V = Q / (4π ε0 R). Hence the capacitance is C = 4π ε0 R. Note also that it takes the form of a length times ε0, as it always must.