1. The problem statement, all variables and given/known data The voltage between the cathode and the screen of a computer monitor is 12 kV. If we assume a speed of zero for an electron as it leaves the cathode, what is its speed just before it hits the screen? a. 8.87*10^7 m/s b. 6.5*10^7 m/s c. 4.2*10^15 m/s d. 7.7* 10^15 m/s e. 5.3*10^7 m/s 2. Relevant equations See below. 3. The attempt at a solution q_e = 1.60*10^-19 C m_e = 9.11*10^-31 kg V = (k*q)/r r = (k*Q)/V = (8.988*10^9(1.60*10^-19)/(12,000 v) = 1.1984*10^-13 m deltaV = (delta U)/q U = deltaV*q = (12,000)*(1.60*10^-19 C) F = E/q = (1.92*10^-15 j0/(1.1984*10^-13 m) = 0.01602 N a = F/m = (0.0160 N)/(9.11*10^-31 kg) = 1.759*10^28 m/s^2 v_f^2 = v_i^2 + 2*a*x v = sqrt[2*1.7587*10^28 m/s^2*(1.1984*10^-13)] = 6.49*10^7 m/s????? 1. The problem statement, all variables and given/known data A surface is so constructed that, at all points on the surface, the E vector points inward. Therefore, it can be said that a.the surface encloses a net positive charge. b.the surface encloses a net negative charge. c.the surface encloses no net charge. d.the surface vector delta S at all points on the surface is necessarily parallel to the electric field vector E. e.the surface vector delta S at all points on the surface is necessarily perpendicular to the electric field E. 2. Relevant equations See below. 3. The attempt at a solution By convention, a negative charge corresponds to an inward electric flux while a positive charge corresponds to an outward flux? Thanks.