A point charge q1 = -2.9 C is located at the origin of a co-ordinate system. Ano

Question

A point charge q1 = -2.9 C is located at the origin of a co-ordinate system. Another point charge q2 = 5 C is located along the x-axis at a distance x2 = 7.8 cm from q1. 1) What is F12,x, the value of the x-component of the force that q1 exerts on q2? 2)Charge q2 is now displaced a distance y2 = 2.6 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 = 6.435 N and points away from q1 and q3. What is the value (sign and magnitude) of the charge q3?

Explanation / Answer

SOLUTION-q1=-2.9C =-2.9x10^-6 C
q2=5C =5x10^-6 C

1)Distance,r=7.8 cm=0.078 m

Electric force between q1 and q2,F=k q1 q2/r^2
=9x10^9x(-2.9x10^-6) x5x10^-6)/(0.078)^2
=-21.44 N

2)Distance,r=(x2^2+y2^2)
=(0.078^2+0.026^2) [y=2.6 cm=0.026 m]
=0.08 m

Now,electric force between q1 and q2,

F=k q1 q2/r^2
=9x10^9x(-2.9x10^-6) x(5x10^-6) /(0.08)^2
=-20.39 N

3)Square of the distance between q2 and q3 is 0.5²* (7.8² +2.6²) = 16.9(cm) ²

= 1.69e-3 m²

The charge q3 must be positive q3, considering the resultant direction .

The force due to q3 on q2 is

K q3*q2/r² = 9e9*5e-6* q3/ 1.69e-3 = 2.67e7 q3

The net force is

- 20.39 + 2.67e7 q3= 6.435

q3 =1.0046e-6 C

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