Two point charges are placed on the x axis. The first charge, q1 = -23.00 nC, is

Question

Two point charges are placed on the x axis. The first charge, q1 = -23.00 nC, is placed a distance
d1 = 13.0 m from the origin along the positive x axis; the second charge, q2 = -14.00 nC, is placed
a distance d2 = 8.00 m from the origin along the negative x axis.
(a) Calculate the electric field (in terms of x and y components) at point A,located at
coordinates (0 m, 7.00 m).
An unknown additional charge q3 is now placed at point B, located at coordinates (0 m , 18.0 m).
(b) Find the magnitude and sign of q3 needed to make the y-­--component of the electric field at point A equal to zero.

Explanation / Answer

Let the 8nC be at the point P and the 6 nC at Q. E1, the field due to q2 at A =(1/4pe ) /d^2 where d = QA = 15m ( sqrt of 12^2+ 9^2) So E1=( 9x10^9 x 6x 10^-9) / 15^2=0.24 N/C along QA Similarly AP = 20 m And the field at A due to q1= 9x10^9 x 8 X10^-9/ 20^2= 0.18N/C along PA The angle between the directions of E1 & E2 = 90 degrees as QAP is a right angled triangle. Therefore the resultant field E is along the Y axis and its magnitude = sqr. root of (E1^2 + E2^2) E=0.30N/Q

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