Answer:
1) y=3x+1
Step-by-step explanation:
passes through the points
(-1.-2) and (3, 10)
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
First you need to find the the mid point of AB. x-coordinate=(-7+8)/2=.5 y-coordinate=(-6-9)/2=-7.5. Find the slope between C and the mid point. m=19.5/1.5=13. Plug into slope intercept form. Y-12=13(x-2).
ANSWER!!!! y=13x-14
