1) 2) Charge q 2 is now displaced a distance y 2 = 2.9 cm in the positive y-dire
ID: 2271195 • Letter: 1
Question
1)
2)
Charge q2 is now displaced a distance y2 = 2.9 cm in the positive y-direction. What is the new value for the x-component of the force that q1 exerts on q2?
3)
A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 5.376 N and points away from q1 and q3. What is the value (sign and magnitude) of the charge q3?
A point charge q1 = -3.4 1/4 C is located at the origin of a co-ordinate system. Another point charge q2 = 5.2 1/4 C is located along the x-axis at a distance x2 = 9.5 cm from q1.Explanation / Answer
the distance = d between q1 and q2 ,
d = sqrt( 0.095 ^ 2 + 0.029^2) = 0.099327 m
distance between q2 and q3 = (d/2) = 0.04966 m
so force on q2 = K * (q1* q2 / d^2) + (K*q2 * q3 / (d/2)^2)
so 5.376 = 9*10^9 * (-3.4 * 10 ^6 * 5.2 * 10 ^-6 / 0.099327^2) + (9*10^9 *5.2 * 10 ^-6 * q3 / 0.04966^2)
solving we get , q3 = 1.1333 *10 ^ -6 = 1.1333 uC
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