PHYS 216 Midterm #1 – Solutions PHYS 216 – Midterm #1 – Solutions and Marking Scheme University of Victoria – Fall 2026 – Steeves Duration: 50 minutes – Instructions: ˆ Answer all 4 problems in an exam booklet (10 marks each, 40 marks total). ˆ You may answer the problems in any order. ˆ Show your work and clearly indicate final answers and units. ˆ Allowed: your formula sheet and a Sharp EL510 calculator. ϵ0 = 8.85 × 10−12 C2 /(N · m2 ), k = 8.99 × 109 N · m2 /C2 , g = 9.81 m/s2 surface area of a sphere = 4πr2 , volume of a sphere = 34 πr3 Integrals (not all are needed; constants of integration omitted): xn+1 Z Z dx xn dx = (n ̸= −1), = ln |x| n+1 x Problem 1. [10 marks] A small sphere of mass m = 1.50 g and charge q = +5.00 × 10−7 C hangs from a light thread between two large, vertical, parallel plates separated by d = 4.00 cm. A potential difference applied across the plates produces a uniform horizontal electric field, and in equilibrium the thread makes an angle θ = 20.0◦ with the vertical, as shown. a. Draw a free-body diagram showing the forces acting on the sphere. b. Find the magnitude of the electric field between the plates and the potential difference across the plates. Which plate is at the higher potential? left plate right plate θ q, m d 1 PHYS 216 Midterm #1 – Solutions Solution. ⃗ horizontal, toward the right a. Three forces: weight m⃗g straight down; electric force F⃗E = q E ⃗ plate; tension T along the thread toward the support. [3] T⃗ θ q, m F⃗E = q E ⃗ y x m⃗g (Arrows drawn to scale: T = 1.57 × 10−2 N, mg = 1.47 × 10−2 N, FE = 5.36 × 10−3 N.) b. Field and potential difference [5] Equilibrium: T cos θ = mg and T sin θ = FE , so FE = mg tan θ. FE = (1.50 × 10−3 )(9.81) tan 20.0◦ = 5.36 × 10−3 N. FE 5.36 × 10−3 E= = = 1.07 × 104 N/C. q 5.00 × 10−7 ∆V = Ed = (1.07 × 104 )(0.0400) = 428 V. Which plate is at the higher potential? [2] The left plate. ⃗ points from the left (Reason, not required: the positive sphere is pushed to the right, so E ⃗ plate to the right plate, and E points from high to low potential.) Problem 2. [10 marks] In a certain region of space the electric potential is V (x, y, z) = 2x2 y − 3yz + 5z, where V is in volts and x, y, z are in metres (the implied constants carry the appropriate units). a. Find the components Ex , Ey , Ez of the electric field and determine the magnitude and direction of the electric field at the origin. b. How much work is required to move a +2.0 µC charge slowly from the origin to (1, 1, 2) m? Solution. ⃗ = −∇V : a. E [6] ∂V ∂V ∂V Ex = − = −4xy, Ey = − = −2x2 + 3z, Ez = − = 3y − 5. ∂x ∂y ∂z ⃗ = −5.0 k̂ V/m: magnitude 5.0 V/m, directed At the origin Ex = 0, Ey = 0, Ez = −5, so E along −z. 2 PHYS 216 Midterm #1 – Solutions b. V (0, 0, 0) = 0 and V (1, 1, 2) = 2 − 6 + 10 = 6.0 V, so ∆V = 6.0 V. [4] W = q ∆V = (2.0 × 10−6 )(6.0) = 1.2 × 10−5 J, independent of the path. Problem 3. [10 marks] A cube of edge length L = 0.20 m has its faces parallel to the coordinate ⃗ = (2.0 ı̂ + 5.0 k̂) × 103 N/C, which lies in the planes, as shown. It sits in a uniform electric field E xz-plane. a. Find the electric flux through the top face of the cube (outward normal n̂ = +k̂) and through the bottom face (outward normal n̂ = −k̂). b. State the net flux through the whole cube, and briefly justify your answer. z ⃗ E top front y x L The uniform field is now switched off and a point charge q = +3.0 nC is placed at the centre of the cube. c. Use Gauss’s law to find the net flux through the whole cube. Then, using symmetry, find the flux through the top face alone. Solution. Each face has area A = L2 = (0.20)2 = 0.040 m2 . For a uniform field through a flat face, Φ=E ⃗ · n̂ A. a. Φtop = E⃗ · (+k̂) A = Ez A = (5.0 × 103 )(0.040) = +2.0 × 102 N m2 /C. [4] Φbottom = E ⃗ · (−k̂) A = −Ez A = −2.0 × 102 N m2 /C (flux enters through the bottom). b. Φnet = 0. [2] There is no charge inside the cube, so by Gauss’s law the net flux is zero. (Equivalently: the top and bottom fluxes cancel, the front and back cancel in the same way, ±Ex A = ±80 N m2 /C, and E⃗ has no y-component, so the two side faces carry no flux.) Either argument is acceptable. q 3.0 × 10−9 c. Φnet = = = 3.4 × 102 N m2 /C (339). [4] ϵ0 8.85 × 10−12 With the charge at the centre the six faces are equivalent, so they share the flux equally: q Φtop = = 56 N m2 /C (56.5). 6ϵ0 3 PHYS 216 Midterm #1 – Solutions Problem 4. [10 marks] A solid non-conducting (insulating) sphere of radius R carries a non-uniform charge density r ρ(r) = ρ0 , r ≤ R, R where ρ0 is a positive constant. There is no charge outside the sphere. a. Show that the total charge on the sphere is Q = πρ0 R3 . b. Use Gauss’s law to find the electric field E(r) inside (r < R) and outside (r > R) the sphere. Solution. a. Divide the sphere into thin shells of volume dV = 4πr2 dr, so dQ = ρ dV . [4] Z R r 4πρ0 R4 Q= ρ0 4πr2 dr = · = πρ0 R3 . 0 R R 4 b. By spherical symmetry E ⃗ is radial and has the same magnitude everywhere on a Gaussian sphere of radius r, so E (4πr2 ) = qenc /ϵ0 . [6] Inside (r < R): only part of the charge is enclosed, r′ Z r πρ0 r4 ρ0 r2 qenc = ρ0 4πr′2 dr′ = , so E= . 0 R R 4ϵ0 R Q ρ0 R 3 Outside (r > R): qenc = Q, so E = = . 4πϵ0 r2 4ϵ0 r2 ⃗ points radially outward. Check: at r = R both expressions give ρ0 R/(4ϵ0 ). In both regions E 4