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
8 = 4w/3 Original Problem.
4w/3 = 8 Flip Flopped the problem around.
4w = 24 Multiplied each side by 3.
w = 6 Divided each side by 4.
Answer: w = 6
Answer:
A
Step-by-step explanation:
A section is decreasing
The answer in is 4.4 bc 1/3 of 13.1(which is half of the total distance) is 4.4
Answer:
12150 ft^3
Step-by-step explanation:
= (45)(20)(10) + (350)(9)
= 9000 + 3150
= 12150 ft^3
