Two charged spheres connected to walls by two springs

[hmparticle9]

Homework Statement

Two spheres of mass $m$ and negligible size are connected to the walls nearest to them by identical springs of force constant $k$. The separation is $a$. When charged to $q$ coulombs each, the separation doubles. Suppose that that the charge on the right is held fixed while that on the left is displaced by a tiny amount $x$ and released. Find the restoring force $F = -k_e x$.

Relevant Equations

$$F_e = \frac{q^2}{4\pi \epsilon_0} \frac{1}{r^2}$$

I have found the spring force constant

$$k= \frac{q^2}{8\pi \epsilon_0} \frac{1}{a^3}$$

but I can not get off the ground with finding $k_e$. Some guidance please?

 

[hmparticle9]

Suppose if we perturb the sphere on the left by a small amount $x$ to the right. Then the sphere on the left will experience a force due to the sphere on the right (pushing the sphere on the left to the left) by

$$F_e = \frac{q^2}{16 \pi \epsilon_0 a^2}.$$

The spring will also exert a force on the sphere on the left given by

$$\frac{q^2}{8 \pi \epsilon_0 a^3}x.$$

The answer is

$$k_e = \frac{3q^2}{16 \pi \epsilon_0 a^3}.$$

I can't add my expressions above, there is a "missing" $a$ in the denominator.

 

[kuruman]

Can you write an expression for the potential energy of the system when one of the spheres is kept in place and the other displaced by $x$?
Note that the system is in equilibrium when the separation between spheres is $2a.$
How is the restoring force near the equilibrium point related to the potential energy?

 

[hmparticle9]

Is my attempt above close or way off? I would like to know why I am incorrect before moving on to another solution method.

No. I cannot write an expression for the PE when one of the spheres is kept in place. Is it something like: $P = \frac{1}{2}k(x-2a)^2$?

 

[kuruman]

Suppose if we perturb the sphere on the left by a small amount $x$ to the right. Then the sphere on the left will experience a force due to the sphere on the right (pushing the sphere on the left to the left) by

$$F_e = \frac{q^2}{16 \pi \epsilon_0 a^2}.$$

Where did you get this expression for the electrical force? If the spheres are initially separated by $2a$ and you bring them closer by $x$, what is the separation? What is the electrical force?

 

[Gavran]

Suppose if we perturb the sphere on the left by a small amount $x$ to the right. Then the sphere on the left will experience a force due to the sphere on the right (pushing the sphere on the left to the left) by

$$F_e = \frac{q^2}{16 \pi \epsilon_0 a^2}.$$

The expression for the electric force is not correct. The explanation is given in the post #5.

The spring will also exert a force on the sphere on the left given by

$$\frac{q^2}{8 \pi \epsilon_0 a^3}x.$$

The expression for the spring force $F_s$ is not correct. The displacement is not $x$. It is $a/2-x$.

The answer is

$$k_e = \frac{3q^2}{16 \pi \epsilon_0 a^3}.$$

The answer is correct.

I can't add my expressions above, there is a "missing" $a$ in the denominator.

The equation $F=-k_ex=F_s-F_e$ can be solved for $k_e$ by simplifying the equation by omitting terms that are significantly smaller than others.

 

[hmparticle9]

Thanks guys :)

Slowly I think I am getting a bit better at physics (very slowly) thanks to you guys.

$$F_e = \frac{q^2}{4 \pi \epsilon_0 (2a - x)^2} = \frac{q^2}{4 \pi \epsilon_0 } \bigg(\frac{1}{4a^2} + \frac{1}{4a^3}x + ...\bigg) \approx \frac{q^2}{16 \pi \epsilon_0 a^2} + \frac{q^2}{16 \pi \epsilon_0 a^3}x$$

$$F_s = \frac{q^2}{8 \pi \epsilon_0 a^3} \bigg( \frac{a}{2} - x\bigg) = \frac{q^2}{16 \pi \epsilon_0 a^2} - \frac{q^2}{8 \pi \epsilon_0 a^3}x$$

subtracting, we obtain the answer. I would like @kuruman to show me the alternative solution involving potential functions please.

 

[kuruman]

I would like @kuruman to show me the alternative solution involving potential functions please.

Write the potential energies with reference to the equilibrium point where the separation is $2a$.
When $x$ is small the spring force is always positive and to the right. Thus, $$U_{\text{sp.}}(x)=-\int_0^x F_{\text{sp.}}dx=-\int_0^x k(\frac{1}{2}a-x)dx=-\frac{1}{2}kax+\frac{1}{2}kx^2.$$The electric potential energy relative to the equilibrium point is $$U_{\text{el.}}(x)=2ka^3\left(\frac{1}{2a-x}-\frac{1}{2a}\right).$$The potential energy function is $U(x)=U_{\text{sp.}}(x)+U_{\text{el.}}(x).$

This function is zero at $x=0$ by construction. Its first derivative is also zero at $x=0$ because we chose $k$ to make it so. The coefficient of the second order (quadratic) term in the series expansion is what we want.

To summarize, this method shifts the zero of potential energy so that, for small $x$ about the equilibrium point, the series expansion for the potential energy looks like a spring-mass system with an effective spring constant, $U(x)\approx \frac{1}{2}k_{\text{eff.}}x^2.$ In other words, the effective spring constant is the second derivative of the potential energy evaluated at the equilibrium point.