2nd semester college physics: "Two identical particles, each with charge 1.25 nC..."?

Question

Two identical particles, each with charge a charge of 1.25 nC, are fixed in the first and fourth quadrants, respectively, of a cartesian coordinate system. Each particle is 0.25 m from the origin along a line that makes an angle of 45 degrees from the axis. (This system can be seen in the figure below.)

A third charge, q', is placed along the y-axis between the other two charges. If the net electric field at the origin is 0, then what is the magnitude of q'?

A third charge, q', is placed along the y-axis between the other two charges. If the net electric field at the origin is 0, then what is the magnitude of q'?

oldschool · Accepted Answer

The vertical components of the fields from the two 1.25nC charges cancel because they are equal in magnitude but in opposing directions. The horizontal components combine:
k*1.25nC*cos(-45)/0.25² + k*1.25nC*cos(+45)/0.25² = +254N/C at 180°
Thus the E field at the origin from a charge at x = 1/√(32) = 254N/C at 0°
kq'/(0.25cos45)² = 254 -----> q' = 254*(0.25cos45)²/k = q' = -884pC

Trending News

2nd semester college physics: "Two identical particles, each with charge 1.25 nC..."?

1 Answer