1) Two charges of q are fixed on the y-axis at y = +a and y = a. A bead of mass

Question

1) Two charges of q are fixed on the y-axis at y = +a and y = a. A bead of mass m with charge +q slides frictionlessly on a horizontal rod positioned along the x-axis.

a) Write the electric force on the bead when it is positioned at any point, x, along the horizontal rod.
b) If the bead is pulled to a position x = A and then let go, describe as completely as possible what its motion will be like at any time, t, after it is let go.
c) Can you write a formula for the position of the bead, x, at any time, t, after it is let go? Take A to be much smaller than a.

Figure 1 – Two charges of q are fixed on the y-axis at y = +a and y = a. A bead of mass m with charge +q slides frictionlessly on a horizontal rod positioned along the x-axis.

Explanation / Answer

(a) let a/x=tan

since the charges are the same at y=+a and y=-a,the net force on the charge +q is - 2*f*cos(- represent the force is attractive and it is towrads origin)

where f= -(1/(4o))*q*q/(a2+x2)   

cos=x/sqrt(a2+x2)

so the electric force acting on the charge +q = -(1/(2**o))*q*q*x/[(a2+x2)3/2]

(b) the bead will have oscillatory motion with the mean position at origin and extreme position at x=+A and x=-A

(c) since the bead undergoes oscillatory motion the equation of motion will be

x=Acos(w*t)

we have accelaration a = d2x/dt2 =-A*w2*cos(w*t) = -w2*x --------eq1

since force from part (a) we have f= -k*q*q*x/a3 where k=1/(2**o) since x<<a we neglect x compared to a

f=m*a

a=-k*q2*x/(a3*m)

from eq1 we have a=-w2*x

so -k*q2*x/(a3*m) =-w2*x

w=sqrt[k/(a*m)]*(q/a)

so the position of bead at time t is x=Acos(w*t) where w=sqrt[k/(a*m)]*(q/a) and k=1/(2**o)

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