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Calculating net electric field?!?
Two point charges lie on the x axis. A charge of +6.34 microC is at the origin, and a charge of -9.80 micro C is at x= 12.0 cm. What is the net electric field at x= -4.14 cm? And x=4.14cm
I have seen similar problems on yahoo answers and I have tried to follow those steps using my numbers and still cannot get the answer correct ... I keep getting 2.99e^7 n/c for -4.14 and that's wrong! Please help!!
1 Answer
- oldschoolLv 78 years agoFavorite Answer
E = kq/r^2
Since the charges are opposite polarities, their fields add between them and subtract outside them.
Positive charges have E fields radially OUT in all directions. Negative charges have E fields radially IN in all directions. Between them their fields are in the same direction. Outside them, their fields are in opposite directions.
E = k*1e-6[ 9.8/0.1614^2-6.34/0.0414^2] = 29.86e6 V/m at x = -0.0414m pointing to the left
Note that 29.86e6 rounds to 29.9e6 = 2.99e7 It would be negative meaning positive magnitude but pointing in the negative x (180 degrees) direction. -2.99e7=2.99e7 at 180 degrees which is correct.
For a point between the charges the fields add:
E = k*1e-6[ 9.8/0.0786^2+6.34/0.0414^2] = 47.50e6 V/m at x = 0.0414m pointing to the right
Note: The distance for the 9.8 charge changes to 0.12-0.0414 = 0.0786
I factored out the 1e-6 to take care of the microC charge