Question

C o webassign.net web/Student/Assignment-Responses/submit?dep=2310631& GVES PRACTICE IT Use the worked example above to help you solve this problem. Charge 4, = 7.50 PC is at the origin, and charge 42 = -5.35 pc is on the r-axis, 0.300 m from the origin (see figure). (a) Find the magnitude and direction of the electric field at point P, which has coordinates (0, 0.400) m. 271412.9695 x magnitude Your response is within 10% of the correct value. This may be due to roundoff error or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. N/C direction 66.6 counterclockwise from the +x-axis (b) Find the force on a charge of 2.00 x 100 c placed at P. 5.42e-3 X Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. N EXERCISE HINTS: GETTING STARTED | I'M STUCK! (a) Place a charge of -10.70 C at point P and find the magnitude and direction of the electric field at the location of 4, due to ,- 7.50 C and the charge at P. magnitude 6e5 N/C 180 X Look at your diagram of the field vectors. is a negative charge and q, is a positive direction charge. What does that tell you about the direction of the electric field vectors E, and E,? Since E is the vector sum of E, and Ey, what range of values is possible for counterclockwise from the + x-axis

I try every example to help me solve it and the computer still says I am wrong even if I am following the equations exactly.

Thank You

Answer 1

Practice It . E py 0.4m= hi 0.3m a ca) ra= √ 0.3² 0.42 = 0.5m a tans (04/03) 2 53.130 Net fieed along & -axis, Ex - Ea coso = x cose C2 8.99x10 x 5.35x10 5.35x 10 cos (53.13) 0.52 - 115431,87 NIC Net field along y -axis, Ey = {. - Ea sino K a2 sing

8.99x19 x 7.5x100 8.99910" X 5.35x10 sir 53,13 0.42 0.52 2 267 497,65 N/ Magnitude - es 2 magnitude - J E2 ² + 8? - 291341 N/a direction, o tani Sx a tan - Lan-1/ 267 497.65 - 115 431.87 ) - 66.66 (6) Force, E = QE = 2x10 291341 - 5.827x10-3 N.

Exeheire E3 - x Net field along X axix, Ex = &, - Ecoso - Ka KV Coso - n? - K Caso V 3 7.5x 10 8 .99x10 x 10.7 x 10 cos53.13 8.99x10" x 0.32 0.52 - 518302.91 a/c Net freed along yaxin, Ey = {z sino - K3 sino He Sin S3,13 = 8,99 10x 10,7x10 - 0.52 = 307817. 19 N/ . magnitude , E = E + 2 = 6.03e5. de tu eftion, otan (Y/Ex) - 30.7°

(b) magnitude of force, F = N E = 5.35 x 106 x 6.03x103 2 3.226 N Direction, o = - 30.7 = 180-20.7= 210.7 cc from ex