Consider 2 parallel plates A and B, separated by d = 10 cm, that form a parallel
ID: 1524174 • Letter: C
Question
Consider 2 parallel plates A and B, separated by d = 10 cm, that form a parallel plate capacitor. A high voltage source is attached across it such that the voltage difference V_AB exists between the A and B plates. Plate B is at ground potential. Each plate has an area of 1.0 m^2. Assume plate A is at x = 0, and plate B is at x = 10 cm. (Plate B has a very small hole in it... so small that its electrical properties are essentially unchanged). The resulting electric field between these two plates is uniform, points to the right, and has a value E = 2.5 times 10^3 V/m (2.5 times 10^5 N/C). a) Consider a test charge of q = +1 mu C starting at x = 10 cm (not touching plate B) at rest. How much change in potential energy is required to push the charge over to plate A (without touching it) at x = 0? a) Consider a test charge of q = +1 mu C starting at x = 10 cm (not touching plate B) at rest. How much change in potential energy is required to push the charge over to plate A (without touching it) at x = 0? b) Knowing that the change in potential, or voltage (Delta V) = Delta PE/q, plot V vs.x. Label the axes with proper units and numerical values. Estimate the slope of the resulting curve (give units). c) Now, an alpha particle (q = +2e, a.m.u. 's = 6.64 times 10^27 kg) is placed at rest a t x = 1.0 cm and accelerates through the capacitor and out the hole in plate B. What is the alpha particle's maximum kinetic energy (in eV)? Make a plot of its Potential Energy and Kinetic Energy (in eV) vs.x. Label the axes with proper units and numerical values.Explanation / Answer
a) change in potential energy = qE * s
= 1 * 10-6 * 2.5 * 105 * 0.1
= 0.025 J
b) slope = 0.025/10-6
= 25000 V
c) maximum kinetic energy = 2.5 * 105 * 0.09 * 2 * 1.6 * 10-19
= 45000 eV