Question

Group Work 2 (E Field Continuous) 1 Given the charge distribution below, what are the components of the E field at point A (black dot) due to q1, q2, and q37 (27050 N/C, 0 deg) 2) What would be the acceleration of a proton placed at point A? (2.59 x 101 m/s') 3) Change the number of charges to 9 equally spaced in 1.0 cm increments. What is the new E field? For 100? For 1000? (1918 N/C; 29600 N/C; 299000 N/C) 11.0 nC (+ 1.0 cm 3.0 cm Y2=1.0 nC @.--- 1.0 cm 31.0 nC+ Step 1: Draw the electric field vectors for each charge at the test charge (much like a free-body diagram) Step 2: Write down your known and unknown variables. Step 3:ill in any missing blanks. Is there any information we need to find beyond what is given? Step 4: Since Electric field is directly related to force, we sum the electric fields as we would forces. (note this answers the second question) Step 5: Answer the questions by solving the summation equations and relating it to the force. Don't forget units Step 6: Use the previous information. How do electric field and force relate? or

Answer 1

Electric field at point A is E_A vector = K q_i (r_A vector - r_i vector)/ Vertical-bar r_A vector - r_i Vertical-bar ^3 E_A vector = Vertical-bar 9 times 10^9 times 1 times 10^-9 (3i^- - j^-)/ Vertical-bar 3i^- - j^- Vertical-bar ^3 times 10^-4 m^2 + 9 times 10^9 times 1 times 10^-9 (3i^-)/ Vertical-bar 3i^- Vertical-bar ^3 times 10^-4 m^2 + 9 times 10^4 times 1 times 10^-9 [3i^- + j^-]/ Vertical-bar 3i^- + j^- Vertical-bar ^3 times 10^4 m^2] E_A vector = 9 times 10^4/(squareroot 10)^3 (3i^- - j^-+ 3i^- + j^-) + 9 times 10^4/9 i^- = 2846.05 times 6i^- + 10000 i^- = 17076.29 i^- + 10000 i^- E_A vector = (27076.29 i^- 0 j^-) N/C ^- E_A vector components are (27050 N/C, 0) Charge on proton = 1.6 times 10^-9 C Mass on proton = 1.67 times 10^-27 kg F_A vector = 1.6 times 10^-19 (27050 i^-) Rightarrow 43280 times 10^-19 i^- N F vector = ma vector ma vector = 43280 times 10^-19 N a vector (1.67 times 10^27) = 43280 times 10^-19 i^-