1)Two charged particles attract each other with a force of magnitude F=2.5 N. If
ID: 1399682 • Letter: 1
Question
1)Two charged particles attract each other with a force of magnitude F=2.5 N. If the distance between the charges is made 3.9 times as large and the charge on one of the particles is made 3.8 times as big, what is the new force?
2)An electric point charge of Q1 = 5.23 nC is placed at the origin of the real axis. Another point charge of Q2 = 3.35 nC is placed at a position of p = 3.98 m on the real axis. At which position can a third point charge of q = -3.57 nC be placed so that the net electrostatic force on it is zero?
3)Let the sign of Q2 be changed from positive to negative. At which position can the point charge q be placed now so that the net electrostatic force on it is zero?
Explanation / Answer
1)
Here ,
initially , let the charges are q1 and q2 and the distance between them is d
electrical force between two charges is given as
F = k*q1*q2/d^2
F = 2.5 N
Now, for changing distance and charge
q1' = 3.8
d' = 3.9 *d
new force = k*q1' * q2/d'^2
new force = k * 3.8 * q1*q2/(3.9*d)^2
new force = 3.8 * 2.5/(3.9)^2
new force = 0.624 N
the new force is 0.624 N
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