The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.2 cm apart with a 26 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential. • Part A What is the electric field strength between the plates? Express your answer to two significant figures and include the appropriate units. ? E- Value Units Submit Reauest Answer • Part B With what speed does an electron exit the electron gun if its entry speed is close to zero? Note: The exit speed is so fast that we really need to use the theory of relativity to compute an accurate value. Your answer to part B is in the right range but a little too big. Express your answer to two significant figures and include the appropriate units. Vexit Value Units

Constants Many chemical reactions release energy. Suppose that at the beginning of a reaction, an electron and proton are separated by 0.110 nm, and their final separation is 0.101 nm. Part A How much electric potential energy was lost in this reaction (in units of eV)? Express your answer using two significant figures. PE eV

Part B The total electron charge within a radius r equals the integral of p(ri) dV from ri = 0 to ri = r. Set up this integral (but don't solve yet), and use it to write an expression for the total charge (including the proton) within a sphere of radius r. > View Available Hint(s) -2r/a047r? dri. the expression for the total charge is The expression for the electron's charge disturbed in the space is -q(r)sphere = Jo Qencl = 0. %3D Пао3 -2r/a04Tr? dri. the expression for the total charge is The expression for the electron's charge disturbed in the space is -q(r)sphere = Jo Qenel = Qproton - q(r)sphere- The expression for the electron's charge disturbed in the space is -q(r)sphere = ( -2r/ao 47TT? dri, the expression for the total charge is Qencl = Qproton - q(r)phere- e-2r/a04rr? drı, the expression for the total charge is The expression for the electron's charge disturbed in the space is -q(r)sphere = Jo - ( Qenel = Qproton +q(r)sphere- Submit Previous Answers Correct

Part A Two 2.0cm x 2.0cm metal electrodes are spaced 1.4 mm apart and connected by wires to the terminals of a 9.0 V battery. What is the charge on each electrode? Express your answer with the appropriate units. • View Available Hint(s) HÁ Value Units Submit ▼ Part B What is the potential difference between electrodes? Express your answer with the appropriate units. • View Available Hint(s) HÀ AV = Value Units