Part I. Answer the following word problems with detailed solution. Use GRESA method in expressing the solution.

For numbers 11 - 15. Complete the table. Charge creating Charge used to Forced Electric field Distance the E field test the E field experienced by test charge intensity (fictional units) 50 nC 1 JC 0.30 N 40 nC 2 JC 3 x 105 N/C 80 nC 1 UC 0.40 N r 60 nC 3 JC 4 x 105 N/C 80 nC 0.60 NSolve the following problems. Show your solution. 1. Two charges of q1=+ 3.0 nC and q2=- 7.0 nC are separated by a distance of 50 cm. What is the electric field strength at a point that is 20cm from q1 and 30cm from q2? The point lies between q1 and q2. 2. Two charges of q1=+ 12.0 nC and q2=- 12.0 nC are 0.100 m apart (such pairs of point charges with equal magnitude and opposite sign are called electric dipoles.). Calculate the electric field caused by q1. 3. Two charges, of q1=+ 3.50 pC and q2=- 4.50 pC are 0.400 m apart. Where along the line joining them will net electric field be zero? Note: E" net= 0 and E* 1 and E 2 are opposite. 4. Four point charges are positioned as such: A(0,1), B(2,1), C(0,0), D(2,0). Charge A = -3 microcoulombs, Charge B = 3 microcoulombs, C = 2 microcoulombs, and D = -2 microcoulombs. What is the net electric field experienced at the origin