The electric field strength at any point from a charged particle is given by E = kq/r^2 and we can use this to calculate the field strength of the two fields individually at the midpoint.
The field strength at midway (r = 0.171/2 = 0.0885 m) for particle 1 is E = (8.99x10^9)(-1* 10^-7)/(0.0885)^2 = -7.041 N/C and the field strength at midway for particle 2 is E = (8.99x10^9)(5.98* 10^-7)/(0.0935)^2 = <span>-7.041 N/C
</span>
Note the sign of the field for particle 1 is negative so this is attractive for a test charge whereas for particle 2 it is positive therefore their equal magnitudes will add to give the magnitude of the net field, 2*<span>7.041 N/C </span>= 14.082 N/C
Answer: $156.86
Step by Step:
136.40 x .15 = 20.46
136.40 + 20.46 = 156.86
Answer:
s> 144 2/3
Step-by-step explanation:
I took the test in k12 and i got it right so yea !
Use socratic to find your answer. :)
The answer for the first one is 1/8 and the answer to the second one is 1/2