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:
I think its 8
Step-by-step explanation:
because if you divided -40 by -5 then its 8
First, you add up the total number of students because we would base our percentage one this.
Total number of students = 1+4+5+10+20+10+5
Total number of students = 55
Now, let's count how many of them got a score of more than 70. For this, you would only count the students belonging to 70-under 80, 80-under 90 and 90-under 100.
Students who got more than 70 = 20+ 10 + 5 = 35
Therefore the percentage is equal to:
Percentage = 35/55 * 100 = 63.64%
The answer is 12.6 that’s the answer right there
