Five charges are evenly spaced on a semi-circle of radius R = 1m. The black dots
ID: 1594266 • Letter: F
Question
Five charges are evenly spaced on a semi-circle of radius R = 1m. The black dots represent the position of charges with Q = +1C and the white dots represent the position of the charges with Q = - square root 2/2 C. The entire semi-circle is tilted 45 degrees with respect to the x axis. And additional particle of charge 1 = -1C is placed at the origin. Your final answer may include constants and trigonometric expressions. Find the magnitude and direction of the electric field at the origin. Find the magnitude and direction of the force on the particle at the origin.Explanation / Answer
(1)First black dot and last black dot will cancel their effect due to same charge and on opposite end.
Now the first white block will produce electric field in positive Y direction
and second white dot will in negative X direction .
Electric field due to the white dot = KQ/r2
Where r is the distnace which is equal to radius R
= K(1/(2)1/2) / R2 = 9*109 / (2)1/2 = 6.364*109 N/C
The magnitude of the both white dot will be same.
But the direction are at 90 to each other
therefore resutant will be = (EX2 + EY2)1/2
= 6.364*(2)1/2*109 = 9*109
This will be in the second quadrant at an angle od 45 from negative X axis
Now due to black dot
Electric field = KQ/R2 = 9*109 *1 / (1)2 = 9*109 Direction will be just opposite that of the white dot field .
Hence net electric field will be ZERO.
Therefore
Force will be ZERO.
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