Problem Solving: Solve and show your solution 1. A uniform electric field of 425 N/C is established between two oppositely charged metal plates. A particle with a charge of + 0.002 C is moved from the bottom (negatively charged) plate to the top plate. (Imagine that the string is tied to the charged that is pulling it upward.) The distance between the plates is 7 am. a. What is the change in the potential energy of the charge? b. What is the change in electric potential from the bottom to the top plate? 2. A 25 HF ( 2.5 x 10 * F)is connected to a source of potential difference of 1000 V. What is the resulting charge on the capacitor?Using the Equation F= k q q:/R?, complete the table below. Show your solution. Electrostatic Electrostatic constant Charge (q,) in Charge (q) in Distance between Force ( F) in (k) in N.my/C coulumb (C) coulumb (C) two charges (R) newton(N) in meter (m) 9.0 x 10 N.m /C 3 x 10 C 7 x10 C 6x 10 m -7.0 x 10 N 9.0 x 10 N.m /C 4 x 10 *C -4.5 x 10 *C -1.0 x 10'N 9.0 x 10 ' N.m /C -3 x 10*C 0.040 m 65 N 9.0 x 10 " N.m /C - 6.0 x 10 - C 0.050 m 9.0 x 10 ' N.m /C2 -2.0 x 10- C 8.0 x 10 * C 0.30 m V. ANALYSIS 1. What will happen to the electrostatic force if the product of their charges is doubled? 2. What will happen to the electrostatic force if the separation of their distance is tripled? 3. What will happen to the distance if the electrostatic force is doubled? 4. What will happen to the electric charge if the electrostatic force increases four times? 5