Question

37- In the diagram below, a 2.0 x 10-6 С charge experiences forces of 3.0 N and 8.0 N at its locatio between charges Q and Q2 Qi 2.0 × 10-1 C F-8.0 N Find the magnitude of the net electric field strength at the location of the 2.0x104 ć charge. A. 2.5x10 N/C B. C. D. 2.8×106 N/C 5.5x106 N/C 1.2x107 N/C

Answer 1

Given, change �q� = 2.0 times 10^- 6 c F_1 = 3.0 N F_2 = 8 N we know that Electric field Strength = force/charge = F_1/q E_1 = F_1/q = 3 N/2 times 10^= -6 c = 1/5 times 10^6 N/c E_1 = 1.5 times 10^6 N/c E_2 = F_2/q = 8 N/2 times 10^- 6 c = 4 times 10^6 N/c E_2 = 4 times 10^6 N/c Now E is the field at a point due to charge �q� and force �F_1� ad �E_2� is th Field at a point due to charge �q� and Force �F_2� the resultant Field when both E = E_1 = E_2 rightarrow E = (1.5 times 10^6) = 4 times 10^6) N/c e = 5.5 times 10^6 N/c Therefore electric field strength ate the location of charge 2.0 times 10^6 c is 5.5 times 10^6 N/c