Exam 1 Solutions. Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1


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1 Exam 1 Solutions Note that there are several variations of some problems, indicated by choices in parentheses. Problem 1 A rod of charge per unit length λ is surrounded by a conducting, concentric cylinder of radius R (see figure) that has no net charge. What is the charge per unit length on the inner (outer) surface of the conductor? (1) λ (2) λ (3) λ / (2 π ε0 r) (4) λ / (2 π ε0 r) (5) 0 The electric field inside of a conductor must be zero (otherwise charges would be accelerating). However, the electric field just outside the charged rod but inside the radius of the conducting shell (r<r) is λ / (2 π ε0 r) by Gauss Law. In order to make the field zero within the volume of the conductor for r>r we must have a charge per unit length of λ on the inner surface of the shell. This charge cannot come for free if the conductor has no net charge, so the charge density on the outer surface is λ. Problem 2 The figure shows two non conducting plastic sheets that are large, parallel, and uniformly charged. The graph below the figure gives the component of the net electric field (in N/C) along an X axis perpendicular through the sheets. What is the charge density of sheet 1 (sheet 2)? (a) 8 ε0 (b) 0 (c) 4 ε0 (d) 2 ε0 (e) ε0
2 From the graph, the electric field is pointing to the negative X direction to the left of both sheets, and to the positive X direction to the right of both sheets. So it appears that the sheets are positively charged and have field directions as indicated by the arrows. The field from an infinitely large charged sheet is E = σ / 2ε0 From the graph we have the following algebraic relations: E 1 î E 2 î = 6î +E 1 î E 2 î = 2î E 1 î + E 2 î = 6î (to the left of both plates) (between plates) (to the right of both plates) The first and third equations are redundant. Solving for the two unknowns, and also using the relation to the charge density gives us: E 1 = 4 σ 1 = 8ε 0 E 2 = 2 σ 2 = 4ε 0 Problem 3 What is the net enclosed charge in the shown cube if the electric field is given by E = ( x 1)î + 4 ĵ (or E = 2( x 1)î + 4 ĵ ) and the cube has a side length of 2? (a) 8 ε0 (b) 0 (c) 4 ε0 (d) 16 ε0 (e) 32 ε0
3 Because of the directions of the electric field, the flux could only be non zero for the top bottom and left right faces. However, the top and bottom faces have a constant field, and so no net flux. The relevant component of the field on the left face is E = î (or E = 2î ), and the field on the right face is E = î (or E = 2î ). In other words, the field points out from the cube in both cases and is considered positive. So the net charge enclosed is: q enc = ε 0 Φ = ε 0 Φ left + ε 0 Φ right = ε 0 E left A left + ε 0 E right A right = ε 0 2(1)(2) 2 = 8ε 0 or = ε 0 2(2)(2) 2 = 16ε 0 Problem 4 The figure indicates an electric field directed only in the + ĵ direction passing through a cube, with the length of the arrows indicating the magnitude of the field at the bottom and top faces. What can be concluded about the net charge contained within the cube? (1) It is positive (2) It is negative (3) It is zero (4) Insufficient information (5) It must be moving or The flux is positive from the top face (field points out of surface), and it is negative for the bottom face (field points into the cube). Since the magnitude is larger for the top (bottom) face, the total net flux is positive (negative). Thus the net enclosed charge by Gauss Law must be positive (negative).
4 Problem 5 A sphere of radius 1 cm contains a positive charge density whose magnitude grows linearly with radius: ρ = ρ 0 r, where ρ 0 = 3 μc/m 4. What is the magnitude of the electric field 0.5 cm from the center of the sphere? (You may need the relation dv = 4πr 2 dr in spherical coordinates.) (1) 2.1 N/C (2) 565 N/C (3) 0 (4) 8.5 N/C (5) 5400 N/C We expect that the electric field will point in the radial direction by symmetry. It s magnitude at a radius r within the sphere can be calculated using Gauss Law: Φ = 4πr 2 E = q enc ε 0 To get the enclosed charge we need to integrate the charge density: q enc = ρ dv = ρ 0 r4πr 2 dr r q enc = 4πρ 0 r 3 dr = πρ 0 r 4 So the field is therefore: 4πr 2 E = πρ 0 r 4 Problem 6 E = πρ 0 r 2 4πε 0 0 ε 0 = 2.1 N/C Four charges are arranged on the vertices of a cube of side length a = 4 cm. The charges are q1 = 100nC, q2 = 200nC, q3 = 300nC and q4 = 400nC. What are the x and y components of the force on q1?
5 (1) Fx = N, Fy = N (2) Fx = N, Fy = N (3) Fx = N, Fy = N (4) Fx = N, Fy = N (5) Fx = N, Fy = N The coordinates at which we observe the field are (x,y) = (0,a). Charge 2 is at (x',y') = (a,a) so its force is F 2 = q 1 q 2 / (4π 0 a 2 ) ( 1,0) Charge 3 is at (x',y') = (0,0) so its force is F 3 = q 1 q 3 / (4π 0 a 2 ) (0,+1) Charge 4 is at (x',y') = (a,0) so its force is 3 F 4 = q 1 q 4 / (4π 0 a 2 2 ) ( 1,1) / 2 Taking into account that q4 = 4q1, q3 = 3q1, and q2 = 2q1, we have F = q 1 2 / (4π 0 ) ( 2 2,3+ 2) (.19N,+.25N) Problem 7 Two charges are held fixed on the x axis. The first charge Q is at x=0 and the other charge 2Q is at x = L. At what value of x can we place an unknown charge so that the force on it from the first two charges vanishes? (1) x = ( 21) L (2) x = ( 2 + 1) L (3) x = 1/3 L (4) x = 1/2 (13 ) L (5) Insufficient information. It's obvious that a point between 0 and L will work. Under this assumption the force on the charge q is qq / (4π 0 ) [+ 1 x 2 2 (L x) 2 ] Setting this to zero requires x = ( 1± 2)L Only the + root corresponds to 0 < x < L, so we must take x = ( 2 1)L
6 Problem 8 Suppose four unknown charges q1, q2, q3 and q4 are placed on the vertices of a square as shown in the figure. If the electric field at the center of the square is 5 N/C ( î + ĵ), which of the following statements can be correct? (1) q1 = q4 and q2 = q3 > 0 (2) q1 = q4 and q2 = q3 < 0 (3) q2 = q3 and q1= q4 > 0 (4) q2 = q3 and q1= q4< 0 (5) None of these statements can be correct Note that q2 = q3 > 0 means that q2 = q3 (opposite charge), and that q2 > 0 and that q3 > 0 (i.e. that q3 < 0) Because the field is perpendicular to the line from q1 to q4, these two charges must be equal so that their fields at the center cancel. Because the field points from q2 to q3, we must have q2 q3 > 0, which is consistent with q2 = q3 > 0. Problem 9 Consider a total charge of 5 pc which is uniformly distributed throughout a sphere of radius 10 cm. What is the magnitude of the electric field at a distance 5 cm from the center of the sphere? (1) 2.25 N/C (2) N/C (3) 18.0 N/C (4) N/C (5) 28.2 N/C
7 By the sphere rule (or Gauss's Law) the field at radius r is the charge enclosed over 4π 0 r 2. The charge enclosed at radius r < 10cm is Q = (r /10cm) 3 5 pc, hence the field has magnitude 5pC r / (4π m 3 ) = 2.25 N/C. Problem 10 Suppose an electric dipole is located at the origin and points in the ĵ direction. What direction does the electric field at (x,y,z) = ( 1 m, 0,0) point? (1) + ĵ (2) ĵ (3) î (4) + î (5) î ĵ Recall that a dipole's field goes up above the dipole axis and then circulates around to come back into the dipole from below. On the plane which is perpendicular to the dipole, the field points in the opposite direction. For this dipole that would be in the + ĵ direction. Problem 11 A linear charge density λ(x) = 4 nc/m x (Note the absolute value function x = Abs (x)) is distributed on the x axis from x=.4 m to to x = +.4 m. What is the electric field vector at the point (x,y,z) = (0,.3m,0)? (1) 29 N/C ĵ (2) 0 (3) 43 N/C ĵ (4) 14 N/C ĵ (5) 85 N/C î By symmetry, the x component of the electric field cancels so only its y component is nonzero. Again, by symmetry, the contribution to this from the region .4m < x < 0 is the same as that from 0 < x < +.4m, so we can just double the integral. If we define D =.3m, L =.4m and a = 4 nc/m^2 then we have L x E y = ad / (2π 0 ) dx 0 3 (x 2 + D 2 2 ) = ad / (2π 0 )[ 1 D 1 ] 29 N/C D L Problem 12 Suppose the electric field is (Ex,Ey,Ez) = (1,2,3) N/C. What is the potential difference in moving from the starting point (x,y,z) = (4,5,6) m to the stopping point (x,y,z) = (7,8,9) N/C? (1)  18 V (2) +18 V (3) +54 V (4)  32 V (5) +50 V
8 The potential difference between any two points is minus the line integral of the electric field. For this constant electric field we have V = E Δ x = [1(7 4) + 2(8 5) + 3(9 6)] volts = 18 volts Problem 13 Suppose the electric potential is V(x,y,z) = a x y 2 z 3, where a is a constant. What is the z component of the electric field? (1) 3 a x y 2 z 2 (2) +3 a x y 2 z 2 (3) a x y 2 z 2 (4) +a x y 2 z 2 (5) a (y 2 z x y z x y 2 z 2 ) Recall that the electric field is minus the gradient of the potential. Hence E z = z axy2 z 3 = 3axy 2 z 2 Problem 14 Consider a negative cylindrically symmetric charge density whose magnitude which grows linearly with the distance s from the central axis. The charge density is ρ(s) = a s, where a = 7 nc/m 4. What is the potential difference from s = 2 cm and s = 5 cm? (1) 10 mv (2) 280 mv (3) 23 mv (4) 15 mv (5) 1.6 mv The linear charge density enclosed within radius s is s λ(s) = 2π ds s a s = 2 3 πas3 0 The electric field at radius s is E(s) = λ(s) / (2π 0 s) = (as 2 ) / (3 0 ) The potential difference from s1 to s2 is minus the line integral s 2 = a / (3 0 ) dss 2 s 1 = a / (9 0 )[(s 2 ) 3 (s 1 ) 3 ] 10 mv.
9 Problem 15 Consider three charges which are arranged at intervals of 60 degrees around a circle of radius R. The first charge is q1 = q. The second charge is q2 = q and lies 60 degrees around in the counter clockwise direction. The third charge is q3 = +q and lies 60 degrees further around in the counter clockwise direction. What is the electric potential energy of this system of three charges in units of q 2 /(4π ε0 R)? (1) 2 + 1/ 3 (2) 2 + 1/ 3 (3) 0 (4) 1/ 3 (5) 1/ 3 Recall that the potential energy of three charges is U = 1 4π 0 [ q 1 q 2 d 12 + q 2 q 3 d 23 + q 3 q 1 d 13 ] Because the three charges are on a circle their distances are d 12 = R, d 23 = R and d 31 = 3R. Hence the energy is q 2 U = [ 1 1+1/ 3] 4π 0 R Problem 16 Consider a charge of 5 pc which is distributed uniformly along the x axis from x= 2 cm to x = 0. If we define the electric potential to vanish at infinity, what is its value on the x axis at x = +3 cm? (1) 1.1 V (2) 2.2 V (3) V (4) 0.8 V (5) V Recall that the potential at position (x,y,z) due to a source dq' at (x',y',z') is 1 dv = d q / (4π 0 ) (x x ) 2 + (y y ) 2 + (z z ) 2
10 For this problem dq' = a/l dx', where a =  5pC and L = 2cm. If we define D = 3cm, the potential is 0 V = a / (4π 0 L) dx / (D x ) = aln(1+ L / D) / (4π 0 L) 1.1 volts Problem 17 L A 1 nf capacitor is formed by two concentric spheres. If the outer sphere has radius 1 m, what is the radius of the inner sphere? (1).9 m (2).8 m (3).7 m (4).6 m (5).5 m Recall that the capacitance of concentric spheres with radii a < b is C = 4π 0 /[1/ a 1/ b] Hence a = [1/ b + 4π 0 / C] 1.9m Problem 18 All three capacitors shown in the figure have the same capacitance C. What is the capacitance of three together? (1) 2/3 C (2) C (3) 3/2 C (4) 3 C (5) 1/3 C C1 and C2 are connected in parallel so their equivalent capacitance is C12 = C1 + C2 = 2C. This capacitor is connected in series with C3, so the total capacitance is [1/ 2C +1/ C] 1 = 2 / 3 C Problem 19 Suppose a 5 F capacitor is charged up using a 3 volt battery, then disconnected after it has been fully charged, and connected to a 4 volt battery. How much additional energy does it draw from the 4 volt battery in becoming fully charged?
11 (1) 17.5 J (2) 22.5 J (3) 40 J (4) 35 J (5).7 J Recall that the energy in a capacitor is U = 1 2 CV 2. So the additional energy to go from 3 volts to 4 volts is 2.5 [169] J = 17.5 J Problem 20 Consider two square parallel conducting plates of side length 2 cm, which hold ±4 nc of charge and are separated by a 3 mm slab of dielectric with κ = 1.5. How much work (in μj) must be done to withdraw the dielectric slab? (1) 2.3 (2) 3.4 (3) 6.8 (4) 0 (5) 4.5 Recall that the capacitance of a parallel plate capacitor with area A and distance d is C = κ 0 A / d. Here κ is the dielectric constant of the material separating the plates. Recall also that the energy of a capacitor with fixed charge Q is U = Q 2 / 2C. Withdrawing the slab changes κ from 3/2 to 1, so the work needed to withdraw the slab is (12/3) the energy for κ = 1, which works out to be about J.
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