Two positive point charges Q are held on the x-axis at x = a and at x = -a

In summary, the conversation revolves around finding the frequency of oscillation for a positive point charge q, released on the x-axis between two other positive point charges Q, using the binomial expansion. The net force on q is calculated and set equal to the equation of motion, and the binomial expansion is used to approximate the result for small distances between q and the origin.
  • #1
S.P.P
39
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I've been doing this question, but can't seem to get past a particular bit. here's the question:

'Two positive point charges Q are held on the x-axis at x = a and at x = -a. A third positive point charge q, of mass m, is placed on the x-axis away from the origin at coordinate x such that |x| << a. The charge q, which is free to move along the x-axis, is then released.

Find the frequency of oscillation of the charge q (hint, use the binomial expansion (1+z)^n = 1 + nz + (n(n+1)Z^2)/2, valid for the case |z| < 1.)

Right, I've no idea where the binomial expansion comes into play, but here's what i got so far:

in SHM F = kx where k is the spring constant. And in electric fields, F = KqQ/r^2 where K = 9*10^9 N.m^2/C^2

if q is placed slightly to the right, of the origin, it will have two forces acting on it. A force from Q nearest to it, and a force from the other Q on the left side of the origin. They will both be repealing it.

The net force at any given position on q would be:

F = KqQ( 1/(a-d)^2 - 1/(a+d)^2 ) where d is the distance of q from the origin. (Call this eqn 1)

Since F = kx where x = d ie F = kd, i should be able to divide eqn 1 by d and have k. But I am pretty sure I've done something wrong up to here, as k is constant for a given system, but varying the length d, gives a different answer of k.

Any help? or any other ways to do this would be very helpful.
 
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  • #2


Originally posted by S.P.P
The net force at any given position on q would be:

F = KqQ( 1/(a-d)^2 - 1/(a+d)^2 ) where d is the distance of q from the origin. (Call this eqn 1)

You're almost there. All you have to do is set F equal to m(d2x/dt2). Of course, you won't be able to solve that exactly, but this is where the binomial expansion comes in handy.

For |d|<<a, we have:

(a-d)-2~1+2d
(a+d)-2~1-2d

Can you try to solve the equation of motion from there?
 
  • #3


First of all, great job on your progress so far! You are correct in your understanding that the charge q will experience two forces from the charges Q, and that the net force can be calculated by subtracting the forces from each other.

Now, let's look at how the binomial expansion comes into play. This expansion is used to approximate the value of a function when the input (in this case, the distance d) is very small. In our case, we are assuming that |x| << a, which means that the distance d is very small compared to the distance a. This is where the binomial expansion will be helpful in simplifying our equation.

Using the binomial expansion, we can rewrite the net force equation as:

F = KqQ( (1/d^2) - (1/(a^2 - 2ad + d^2)) )

= KqQ( (1/d^2) - (1/a^2) + (2ad/a^4) - (d^2/a^4) )

= KqQ( (1/d^2) - (1/a^2) + (2d/a^3) - (d^2/a^4) ) (using the hint given)

Now, we can see that the term (1/a^2) is much larger than the other two terms, since we are assuming that d is very small. Therefore, we can ignore the other two terms, leading to the simplified equation:

F = KqQ(1/a^2)

= kx (since x = d)

Therefore, we can see that the spring constant k is equal to KqQ/a^2. This means that the frequency of oscillation can be calculated using the equation:

f = 1/(2π) * √(k/m)

= 1/(2π) * √(KqQ/(ma^2))

= 1/(2π) * √(9*10^9 * q * Q/(ma^2))

I hope this helps you understand how the binomial expansion is used in this problem. Keep up the good work!
 

1. What is the electric field at the origin due to these two point charges?

The electric field at the origin is zero because the two point charges have equal magnitudes and opposite signs, canceling each other's electric field.

2. How does the distance between the two point charges affect the electric field at a point on the x-axis?

The electric field at a point on the x-axis is inversely proportional to the square of the distance between the two point charges. As the distance increases, the electric field decreases.

3. Can the two point charges have the same sign and still create a non-zero electric field at the origin?

Yes, the two point charges can have the same sign and still create a non-zero electric field at the origin. The electric field would be the sum of the individual electric fields created by each point charge, resulting in a non-zero net electric field at the origin.

4. How does the magnitude of the point charges affect the electric field at a point on the x-axis?

The electric field at a point on the x-axis is directly proportional to the magnitude of the point charges. As the magnitude increases, the electric field also increases.

5. What is the direction of the electric field at a point between the two point charges?

The direction of the electric field at a point between the two point charges depends on the relative distances from each point charge. If the point is closer to one point charge, the electric field will be directed towards that charge. If the point is equidistant from both point charges, the electric field will be directed away from the midpoint between the two charges.

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