Charges and Coordinates The charges andcoordinates of two charged particles held

Question

Charges and Coordinates The charges andcoordinates of two charged particles held fixed in the xyplane are qA = +3.3 µC, xA = 3.5 cm,yA = 0.50 cm, and qB = -4.0µC, xB = -2.0 cm, yB= 1.5 cm. (a) Find the magnitude and direction of theelectrostatic force on qB.
N
° (counterclockwise from the +xaxis)

(b) Where could you locate a third charge qC =+4.4 µC such that the netelectrostatic force on qB is zero?
xC = cm
yC = cm (a) Find the magnitude and direction of theelectrostatic force on qB.
N
° (counterclockwise from the +xaxis)

(b) Where could you locate a third charge qC =+4.4 µC such that the netelectrostatic force on qB is zero?
xC = cm
yC = cm

Explanation / Answer

F = -qa*qb/(k*r1^2) r1^2 = x^2 +y^2 x = 3.5+2= 5.5 y = 1.5-.5 = 1 ===> r^2 = 5.5^2 +1 That should get you the magnitude of the force. To get the direction use arcsin(y/r) = theta. In this case,your y = -1. For part b, now that you know the force on Qb, just set F = to thenegative of the force that you solved in part A. so essentially you're saying: qa*qb/(k*r1^2) = -qc*qb/(k*r2^2) Here, you're solving for r2. That gets you the magnitude of thedistance vector. Now, you know that charge qc should lie on thesame line as the one that connects qa and qb. So, the angle will be180 degrees opposite your answer for the first part. So, using the theta found in the first part: yc = sin(theta +180)*r2 + 1.5 xc = cos(theta +180)*r2 -2

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