charge "Q1" = + 6 nC is on y axis at y = 3cm and Q2 = -6 nC is on y axis at y = -3 cm .calculate force on a test charge q = 2 nC placed on x -axis at x=4 cm .
Asked by Zubin Khalfay | 23rd Jul, 2013, 12:28: PM
let Q1 = +6 nC be placed at point A (y = +3)
Q2 = -6nC be placed at point B (y = -3)
test charge q is placed at C(4,0)
Now, Force due to charge Q1 = k*6*2/root(3^2 + 4^2) along AC = 12k/5 along AC
Force due to charge Q2 = k*6*2/root(3^2 + 4^2) along CB= 12k/5 along CB
Hence, now, if you will resolve the forces, their horizontal components will get cancelled, but the vertical components will add up in the -y direction.
hence, net force = 2*12k/5 * 3/5 (-j)
Net force = 36k/25 (-j) where k is the Coulomb's constant.
Answered by | 26th Jul, 2013, 05:39: AM
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