Two identical particles, each having charge +q, are fixed in space and separated by a distance d.

+5 votes

Two identical particles, each having charge +q, are fixed in space and separated by a distance d. A third particle with charge −Q is free to move and lies initially at rest on the perpendicular bisector of the two fixed charges a distance x from the midpoint between those charges (see figure below).

image

(a) Show that if x is small compared with d, the motion of −Q is simple harmonic along the perpendicular bisector. (Do this on paper. Your instructor may ask you to turn in this work.)

(b) Determine the period of that motion. (Use the following as necessary: π, q, Q, m for the mass of charge Q, d, and ke.)
T =

(c) How fast will the charge −Q be moving when it is at the midpoint between the two fixed charges if initially it is released at a distance a << d from the midpoint? (Use the following as necessary: π, q, Q, m for the mass of charge Q, d, and ke.)
v =

asked Jan 15, 2012 in Physics by Joey33 ~Expert~ (1,216 points)
    

1 Answer

+3 votes

First, recall that spring force is F = -kx

The top charge exerts a force on the negative charge (ke*qQ) / ((d/2)^2 + x^2) which is directed upwards and left, at an angle theta.   Now notice that the y-components of the charges canel each other out so there is no y-component in the final result.

So procceed in finding the x-component of the 2 forces.  Rather than doing this twice, because of the symmetry you just need to mulitply the answer by 2.

The total force is: -2(ke*qQ) / ((d/2)^2 + x^2)^(1/2) * cos(theta) i

Now you must solve for cos(theta) in terms of the geometry of the figure.

 

Now to find the period.  Recall that the formula

acceleration = angular frequency^2 * x

Cos = adjacent/ hypotenuse, so:

cos = x / ((d/2)^2 + x^2)^(1/2)

Now plug back into the equation for total force to get :

[-2(ke*qQ) / ((d/2)^2 + x^2)^(1/2)] * [x / ((d/2)^2 + x^2)^(1/2)] i = simply to get:

[-2(ke*qQ) * x  / ((d/2)^2 + x^2)^(3/2)] i

In the question it states that x is small, so we can safely remove x from the denominator and simplify resulting in:

[-2(ke*qQ) * x  / ((d/2)^2 )^(3/2)] i

Notice how the first part of this is a constant? easier to see when you rewrite it like so:

F = - [2(ke*qQ) / ((d/2)^3] * x,

(the first part in brackets is the k-value or the spring constant.)

now recall that F = ma, and hence to get acceleration just divide by m, and so:

a = - [2(ke*qQ) / m*((d/2)^3] * x

Now use the formula:  a = - (angular frequency) ^2 * x

and solve for angular frequency.

Now recall that the Period is simply:  T = 2pi / (angular frequency)

and so:

T = (pi/2) [(m*d^3)/(ke*qQ)]^(1/2)

 

and vmax is just angular frequency times acceleration:

so vmax = 4a * [(ke*qQ) / (m*d^3)] ^ (1/2)

answered Jan 19, 2012 by pokemonmaster ~Expert~ (3,856 points)
edited Jan 19, 2012 by pokemonmaster



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