In the figure below, particles 2 and 4, of charge -e, are fixed in place on a y

Question

In the figure below, particles 2 and 4, of charge -e, are fixed in place on a y axis, at y_2 = -10.0 cm and y_4 = 5.00 cm. Particles 1 and 3, of change -e, can be moved along the x axis. Particle 5, of charge +e, is fixed at the origin. Initially particle 1 is at x_1 = -15.0 cm and particle 3 is at x_3 = 15.0 cm. (a) To what x value must particle l be moved to rotate the direction of the net electric force F_net vector on particle 5 by 30 degree counterclockwise? (Assume particle 5 is located at the origin of the system.) -3.66 m (b) With particle 1 fixed at its rev. position, to what x value must you move the particle 3 to rotate F_net vector back to its original direction? +3.66 m

Explanation / Answer

(a)

Initially the field at origin is,

E1 = (1/40)e/(5cm)2 - (1/40)e/(10cm)2 = 300(1/40)e along positive y-axis.

So, initial value of force on 5 is,

Fnet1 = 300(1/40)e2 along +y-axis

Now let's suppose charge 3 is moved to a position which at a distance d from origin on x-axis and on left of origin.

Now, x-component of field at origin is,

Ex = (1/40)e/(d)2 - (1/40)e/(0.15m)2towards negative x-axis.

Fx = (1/40)e2/(d)2 - (1/40)e2/(0.15m)2 towards negative x-axis

Fy = 300(1/40)e2 along +y-axis

tan300 = Fx/Fy = [ (1/40)e2/(d)2 - (1/40)e2/(0.15m)2]/[300(1/40)e2]

or, d = -0.0677 m or  -6.77 cm

So particle 1 should be moved to x = -0.0677 m or x = -6.77 cm

***************************************************************************************************

(b)

Particle 3 should be moved to x = +0.0677 m or x = +6.77 cm

***************************************************************************************************
This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification, modification or correction, feel free to ask.....

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---