SS-I Physics 0175 Recitation 2 (5/22/18) Name (all group members): 1. The figure shows a cross section through a very large nonconducting slab of thickness d = 9.40 mm and uniform volume charge density ρ = 5.80 fC/m3. The origin of an x axis is at the slab’s center. What is the magnitude of the slab’s electric field at an x coordinate of (a) 0, (b) 2.00 mm, (c) 4.70 mm, and (d) 26.0 mm? Hints: It’s similar to an infinite sheet of charge but has thickness and therefore, the enclosed charge is different in (a), (b), (c), (d). Ans: We use a Gaussian surface in the form of a box with rectangular sides. The cross section is shown with dashed lines in the diagram to the right. It is centered at the central plane of the slab, so the left and right faces are each a distance x from the central plane. We take the thickness of the rectangular solid to be a, the same as its length, so the left and right faces are squares. The electric field is normal to the left and right faces and is uniform over them. Since ρ = 5.80 fC/m3 is positive, it points outward at both faces: toward the left at the left face and toward the right at the right face. Furthermore, the magnitude is the same at both faces. The electric flux through each of these faces is Ea2. The field is parallel to the other faces of the Gaussian surface and the flux through them is zero. The total flux through the Gaussian surface is 22 .EaΦ = The volume enclosed by the Gaussian surface is 2a2x and the charge contained within it is 22q a xρ= . Gauss’ law yields 2ε0Ea2 = 2a2xρ. We solve for the magnitude of the electric field: 0/ .E xρ ε= (a) For x = 0, E = 0. (b) For x = 2.00 mm = 2.00 × 10−3 m, 15 3 3612 2 20(5.80 10 C/m )(2.00 10 m)1.31 10 N/C.8.85 10 C /N mxEρε− −−−× ×= = = ×× ⋅ (c) For x = d/2 = 4.70 mm = 4.70 × 10−3 m, 15 3 3612 2 20(5.80 10 C/m )(4.70 10 m)3.08 10 N/C.8.85 10 C /N mxEρε− −−−× ×= = = ×× ⋅ (d) For x = 26.0 mm = 2.60 × 10−2 m, we take a Gaussian surface of the same shape and orientation, but with x > d/2, so the left and right faces are outside the slab. The total flux through the surface is again 22EaΦ = but the charge enclosed is now q = a2dρ. Gauss’ law yields 2ε0Ea2 = a2dρ, so 15 3 3612 2 20(5.80 10 C/m )(9.40 10 m)3.08 10 N/C.2 2(8.85 10 C /N m )dEρε− −−−× ×= = = ×× ⋅ 2. An infinite nonconducting sheet has a surface charge density σ = +5.80 pC/m2. (a) How much work is done by the electric field due to the sheet if a particle of charge q = +1.60 × 10−19 C is moved from the sheet to a point P at distance d = 3.56 cm from the sheet? (b) If the electric potential V is defined to be zero on the sheet, what is V at P? Hints: Express V as an integral of E. However, no integration is needed since E is constant. Ans: (a) The work done by the electric field is19 12 20 0012 2 200 021(1.60 10 C)(5.80 10 C/m )(0.0356 m)2 2 2(8.85 10 C /N m )1.87 10 J.f diq q dW q E ds dzσ σε ε− −−−× ×= ⋅ = = =× ⋅= ×∫ ∫ (b) Since V – V0 = –W/q0 = –σz/2ε0, with V0 set to be zero on the sheet, the electric potential at P is 12 2212 2 20(5.80 10 C/m )(0.0356 m)1.17 10 V.2 2(8.85 10 C /N m )zVσε−−−×= − = − = − ×× ⋅ 3. A uniform electric field E of magnitude 2.00 × 103 N/C has been set up between two horizontal plates. L = 10.0 cm and d = 2.00 cm. An electron is shot from the left edge of the lower plate with initial velocity 0v that makes an angle θ = 45.0° with the lower plate and has a magnitude of 6.00 × 106 m/s. (a) Will the electron strike any of the plates? If so, which one? Hints: Compare with a projectile to use corresponding equations. Ans: (a) The electric field is upward in the diagram and the charge is negative, so the force of the field on it is downward. The magnitude of the acceleration is a = eE/m, where E is the magnitude of the field and m is the mass of the electron. Its numerical value is a =× ××= ×−−160 109 11 10351 10193114... .C 2.00 10 N Ckgm s32chch We put the origin of a coordinate system at the initial position of the electron. We take the x axis to be horizontal and positive to the right; take the y axis to be vertical and positive toward the top of the page. Then, ()( )26 22 220max2146.00 10 m s sin 45sin2.56 10 m.22 3.51 10 m svyaθ−× °= = = ×× Since this is greater than d = 2.00 cm, the electron might hit the upper plate (it will hit the upper plate if the corresponding x coordinate is less than L.) Check whether x coordinate is less than L: We find the x coordinate of the position of the electron when y = d. Since v06 66 00 10 4 24 10sin m s sin 45 m sθ= × °= ×. .ch And 2 2 351 10 0 0200 140 1014 132ad = × = ×. . .m s m m s22dibg the solution to d v t at= −0122sinθ is ( )226 6 13 22 20 02149(4.24 10 m s) 4.24 10 m s 1.40 10 m ssin sin 23.51 10 m s6.43 10 s.v v adtaθ θ−× − × − ×− −= =×= × The negative root was used because we want the earliest time for which y = d. The x coordinate is ()()6 9 20cos 6.00 10 m s 6.43 10 s cos45 2.72 10 m.x v t− −= = × × ° = ×θ This is less than L so the electron hits the upper plate at x = 2.72