Two rods A and B of lengths L and L/2, respectively, are arranged as shown in th

Question

Two rods A and B of lengths L and L/2, respectively, are arranged as shown in the diagram below where d L/2. Here, L 2.40 m and the charge on rod A is 55.0 HC. If the net electric field at the midpoint between the two rods is zero, what is the charge (magnitude and sign) on rod B? The field at a distance y on the perpendicular bisector of a rod of length L is given by Erod magnitude How should the magnitudes of the two fields be related for the net field to be zero at the location P? What direction of the field due to rod B will result in a zero net field at P? HC sign positive

Explanation / Answer

given
rod on left
l1 = L
q1 = 55 uC

rod on rith
l2 = L/2
q2 = ?

net field at mid point is 0
hence
charge q1 and q2 have same sign
hence

2*kq1*(L/2)/sqrt(L^2/4 + d^2/4) = 2k*q2*(L/4)/sqrt(L^2/16 + d^2/4)
d = L/2
q1/sqrt( 5) = q2/2*sqrt( 2)
q2 = 0.632455532*q1 = 34.785054 uC

signs is ofcourse +ve

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