1. For each charge, you need to do a vector addition of the coulomb forces due to the two other charges. Because of symmetry, each corner charge experiences a force away from the triangle, along a line perpedicular to the opposite side.
2. The field equals the force divided by the proton charge, 1.6*10^-19 C
The field points south.. the same as the force
1. Three particles with charges of +11 mC each are placed at the vertices of an equilateral triangle with sides of 15.0 cm. What is the magnitude and direction of the net force on each particle?
2. A proton is released in an uniform electric field and it experiences a force of 3.2 x 10-14 N toward the south. What are the magnitude and direction of the electric field?
2 answers
1)magnitude of force is 4,84,00,000 N.along a straight line which makes 30degrees with the forces
2)magnitude of force is 1.6*10-14
2)magnitude of force is 1.6*10-14