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
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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:
The first one is 36 i think and the second is 28 i think
Step-by-step explanation:
Answer: You can multiply the top equation by -1 to eliminate the x variable.
And the solution is (2,4/3) in case you need it.
Step-by-step explanation:
2x + 3y = 8
2x + 6y = 12
If you multiply the upper equation or down equation by one, you will be able to eliminate the x variable.
-1( 2x + 3y) = -1(8) New equation: -2x -3y = -8.
Add the new equation you got by multiplying the top equation by -1 to the bottom equation.
Add them: -2x -3y = -8
2x + 6y = 12
3y = 4
y = 4/3
You can now input the value for y into the one of the equations and solve for x.
-2x - 3(4/3) = -8
-2x -4 = -8
+4 +4
-2x = -4
x = 2
babe what are you doing on brainly?
