Answer:
2
Step-by-step explanation:
it would have to be a, because using pythagorean theorem, we can find that a squared + b squared = c squared. substitute that for 1 + b squared = 4; b squared=3 b = square root of 3
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:
2/3x
Step-by-step explanation:
This is a simple ax graph. all we need is the slope. the slope is the amount the y goes divided by the amount the x goes. essentially
change in y / change in x
2/3
2/3x
