Question

1. The objectives of this lab are (1) to determine initial mechanical energy of a charge (2) to determine final mechanical energy of a charge (3) to prove the law of conservation of energy for the electric charge II. Apparatus(Equipment): Voltage source, measuring tape, smart timer. III. Procedure: A charged body having charge Q = 0.002 C, mass M= 0.004 Kg, was initially held at rest at a voltage line located at height=H=D= 2m. Varying amounts of voltages were applied at the voltage line. The time the charged body took to fall to the zero-voltage level of the ground were measured using a digital smart timer. Attracted by the electrical force, the charge falls to the zero-voltage ground in the indicated times. Since the electrical forces on the charge are stronger than gravity, gravitational forces are negligible. IV. Raw Data: The voltages applied, and the corresponding falling times of the charged particle are indicated in the raw data table below Voltage (V) Time (s) Initial Mech. Energy (E) (Joules) Acceleration(m/s) Final Mech. Energy (Ed) (Joules) 200 0.280 400 0.191 600 0.165 800 0.149 1000 0.132 V. DATA ANALYSIS: (1) For each row, showing your steps on a separate attached paper, calculate the initial mechanical energy (El) of the charge, the acceleration of the charge and the final mechanical energy (El) of the charge. Fill in your answers in the table. (2) For each row, calculate the percent difference between initial mechanical energy (EI) and final mechanical energy (EF). INITIAL ENERGY-FINAL ENERGY Percent difference = INITIAL ENERGY+FINAL ENERGY X 100 % 2

Answer 1

A charge body having 0.002c) e Mass (M= 0.004 kg was divitially at fest' at a Voltage line located at height H=D=2m. Varying amount of voltages are applied at voltage line. The time the charged body took to fall at 36-voltage level of ground were measured Potential solution: :: Work done in moving a charge a from from VA to VB will stored as Mechanical energy charge particle Initial mechanical energy = @ (VA-VE) Initial mechanical VA line 2m. VB zelo potential of the level W where UB=0 uergy = QUA Electrical force q -QË - - Q/-ay da f -QdV F = Q(VA-Ve) (d-X=2m VB-0 f=ma QUA -mxa QXVA accelerations 2 QM Suice Eleclical force is conser ervative force and acceleraton is constant we can apply s-ut that 2

also we can use V=utat V=at fuial Mechanical Energy = 1 m 2 £m(at) 9 Since we have derived all the formula , we will explani the calculation for 1 valve and then simply fill the table (3) W=2008 t=0-280, 0.002 x200 QM ។ Initial Mech Energy 0.002 / 200 (i Qva) -0.40 Aseeretat, Acceleration = 240.004 = 200 50m/s2 fuial Mechanical Energy Făm lat)2 gewen t= 0.280 ==x0.004 ( 508 0.280) 2 = 0-392 Percent dift Initial E- finialt. X100 = 0.4-0-312 x 100 Ini Et Final E 0.4+0.392 *49 = 0.008 X100 % 1. Percent ditt = 2% 0.396

Voltage Time Initial Mech, Energy Acceleration Final Mechanical Percentage different ms) (v/ ( s) - QXVA Energy 2M mlat)2 0.002 VA 2x0.004 VA/4 = V/4 0.280 0.4T 0.392 J 1.) 200 27 0.191 0.002 x 400 9-27 2) 400 =0.8T 50m/s2 - 46014 50-7296J = 100m/32 О. - 1.225 660/4 = 2 150 m/s2 860/4 = goom154 2.06% (18) 600 0.165 1=0.002 x 600 =12 J 1-7765 10% (1 = 1.6 J (4) 800 0.149 = 2,178 8.5% 250 m/s2 0.132 =QJ 5) 1000