Electric Force Three positive charges with values 2q 2q, q. and q are arranged a
ID: 2002855 • Letter: E
Question
Electric Force Three positive charges with values 2q 2q, q. and q are arranged along the x axis, as shown in the figure. The spacing between the charges is J. 1 d 2 d The charges are labeled 1,2, and 3 for convenience, as indicated. What is the direction of the net force on the charge 2? Why? What is the magnitude of the net force on the charge 2? Suppose a fourth charge of value q 12 is added to the system at a location which makes the net force on charge 2 equal to zero. Where should the fourth charge be placed? Next, assume that the fourth charge is removed and we have the original three charges. Suppose that all 3 charges are suddenly released from the positions shown in the figure. Assuming that all charges have the same mass m , find the acceleration of each of the 3 charges. (Give the magnitude and direction.) Now suppose the two charges at the ends are fixed in place at the positions shown in the figure, but the middle charge is allowed to move. There is a location where the net force on the middle charge is zero. Set up an equation which gives the condition for zero net force on the middle charge. (You don't need to solve the equation, just set it up.)Explanation / Answer
A) since all changes are positive they exerts repulsive forces .
Force F = k q1q2 / r^2
It varies directly with charge so force due to charge 1 on 2 is double to the force by 3 on 2.
The net force is directed towards charge 3 on the line joining them.
B) Force due to 1on 2 F1 = k *q*2q /d^2
Force due to 3 on 2 F 3 = k *q*q /d^2
Net force F = F1 - F3 = k*q^2 /d^2 (2-1)
= k*q^2 /d ^2
C) charge 4 = q /2
Let the distance of 4 from 2 = a
It should be placed on line joining 2 and 3 to counter the net force
To make the resultant force zero
Force due to charge 4 ( F4) = net force F
k*q*q /2 *a^2 = kq^2 /d^2
a^2 = d^2 /2
a = d /(2)^1/2
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