A positive test charge of 9.0E-5 C is placed in an electric field of 35.0 N/C intensity. What is the strength of the force exerted on the test charge?© BrainMass Inc. brainmass.com August 14, 2018, 9:16 am ad1c9bdddf
q = 9.0 * 10^ -5 C (This has 5 significant figures), E = 35.0 N/C (This has 3 ...
Force in an electric field is determined. The test charge exerted is determined.