# Thread: [SOLVED] Electrostatics and Mechanics?

1. ## [SOLVED] Electrostatics and Mechanics?

In the diagram shown $R$ is a long smooth fixed non-conducting rod of uniform linear charge density $\lambda$ and radius of cross-section $r$. $A$ is a non-conducting uncharged smooth ring of mass $m$ which fits completely on the rod and can move horizontally on the rod. $B$ is a non-conducting charged small particle of mass $2m$ and charge $Q$ and connected to $A$ by an inextensible string of length $l$. $B$ is released from rest from the position shown in the diagram. Find velocity of $B$ and tension in the string when string becomes vertical. Assume gravitational acceleration $g$ and $\lambda Q > 0$.

2. Originally Posted by fardeen_gen
In the diagram shown $R$ is a long smooth fixed non-conducting rod of uniform linear charge density $\lambda$ and radius of cross-section $r$. $A$ is a non-conducting uncharged smooth ring of mass $m$ which fits completely on the rod and can move horizontally on the rod. $B$ is a non-conducting charged small particle of mass $2m$ and charge $Q$ and connected to $A$ by an inextensible string of length $l$. $B$ is released from rest from the position shown in the diagram. Find velocity of $B$ and tension in the string when string becomes vertical. Assume gravitational acceleration $g$ and $\lambda Q > 0$.
Making the assumption that the particle is not going to be bound to the rod at the outset (that is to say the electrostatic force in the beginning isn't greater than the weight of the particle) then this problem is more or less a simple work-energy problem. The difficult bit will be calculating the potential energy function for the rod. I'm assuming you have this. So then, since the electric force is conservative we have
$\Delta E = \Delta K + \Delta U + \Delta P = 0$
where U is the gravitational potential energy.

This equation will give you the speed of the particle at it's lowest point on the trajectory and you can finish the problem by noting that the net force on the particle is equal to the centripetal force on it. That an Newton's 2nd Law will give you the tension.

If you need more details than this, simply ask.

-Dan

3. Originally Posted by topsquark
Making the assumption that the particle is not going to be bound to the rod at the outset (that is to say the electrostatic force in the beginning isn't greater than the weight of the particle) then this problem is more or less a simple work-energy problem. The difficult bit will be calculating the potential energy function for the rod. I'm assuming you have this. So then, since the electric force is conservative we have
$\Delta E = \Delta K + \Delta U + \Delta P = 0$
where U is the gravitational potential energy.

This equation will give you the speed of the particle at it's lowest point on the trajectory and you can finish the problem by noting that the net force on the particle is equal to the centripetal force on it. That an Newton's 2nd Law will give you the tension.

If you need more details than this, simply ask.

-Dan
Such an assumption is not required as the force is repulsive since $Q \lambda > 0$ which can only be true if both charges are positive or negative.

Also the string is not attached to a fixed point. It is attached to a ring which can move along the rod.

So this must be considered as a regular mechanics problem but just with an additional force - not that this invalidates energy arguments.

Physically what would happen is that the rod repels the mass away from it but as it is attached to the string it pulls the ring along the rod which in turn causes the mass to swing until there's a point where the mass and ring are vertically in line but have opposite velocities at which point I believe they'll carry out motion similar to a pendulum but with the ring moving side to side in tandem with the mass (but in opposite directions).

Fardeen, I'll put up a solution when I get a chance and I'll send you that list when I get a chance to have a look through the books on my bookcases.

4. ## The solution

So here's the solution I promised - sorry it took me a while to get back to you.

Spoiler:

The way to do this is first to deal with the electrostatics of the problem i.e. figure out the nature of the repulsive force due to the charges. Based on the nature of this force we can then analyse the mechanics of the problem.

Electrostatics

Given the fact that this is a long rod we can assume that its length is much greater than $l$ or $r$ and so we can take it to be effectively infinite. So assume the rod is of infinite length.

Note that the question states that $\lambda$ is the linear charge density which means it is the charge per unit length of the rod.

Consequently by symmetry arguments you should be able to see that the electrostatic repulsive force must be perpendicular to the length of the rod and so acting radially outwards.

To work out the force on $B$ I used Gauss's Theorem (otherwise known as a special case of Stoke's Theorem) with Maxwell's law of electromagnetism regarding divergence of electric field as it's the easiest way of working it out for me. Thus the electrostatic repulsive force, $F_E$ , is given by

$\color[rgb]{0,0,1}\boxed{F_E = \frac{Q \lambda }{2 \pi \epsilon_0 \rho}}$

where $\rho$ is the distance from the rod's center radially outwards to B.

However I know that you are only 16 and so this is perhaps too advanced a technique so alternatively the force can be found by integration of the force from Coulomb's law as follows. To do this we consider cylindrical slabs of the rod of length $\mathrm{d}z$ ( $z$-axis along the axis of the rod) which carry charge $\lambda \mathrm{d}z$. For convenience we have the origin of the $z$-axis at the centre of the rod directly above $B$ at its initial position. As I stated earlier from symmetry arguments we only need to integrate over the component of the force which is perpendicular to the z-axis i.e. radially outwards from the rod. Hence the force due to the slab from Coulomb's law is

$dF_E = \frac{\rho}{\sqrt{\rho^2 +z^2}} \, \frac{Q \lambda }{4 \pi \epsilon_0 \left( \rho^2 +z^2 \right) } \, \mathrm{d}z$,

$\quad = \frac{Q \lambda \, \rho}{4 \pi \epsilon_0 \left( \rho^2 +z^2 \right)^{\frac{3}{2}} } \, \mathrm{d}z$,

and integrating we find

$F_E = \int_{-\infty}^{\infty} \frac{Q \lambda \, \rho}{4 \pi \epsilon_0 \left( \rho^2 +z^2 \right)^{\frac{3}{2}} } \, \mathrm{d}z$.

Using the substitution $\rho \tan \theta = z$ we have

$F_E = \frac{Q \lambda}{4 \pi \epsilon_0 \rho } \int_{-\pi/2}^{\pi/ 2} \cos \theta \, \mathrm{d}\theta = \frac{Q \lambda}{2 \pi \epsilon_0 \rho }$

as stated earlier.

Mechanics

Now we know that the electrostatic force is in the same direction as gravity so we can treat this in much the same way as we would if there was only a gravitational force.

Ordinarily if the particle was anchored to the rod then the velocity would be given by the kinetic energy of $B$ being equal to the potential energy lost. However, for this problem $B$ is not anchored and so the kinetic energy gained must be shared between the ring $A$ and particle $B$. To determine in what proportion the kinetic energy is shared between the two note that the net force acting on $B$ (due to gravity and electrostatic repulsion) only ever acts vertically down and so the system of $A$ and $B$ cannot have any change in their lateral (parallel to the rod) momentum. In other words the momentum of $A$ and $B$ parallel to the $z$-axis must be equal and opposite.

This means that if $A$ has speed $v_A$ and $B$ has lateral speed $v_B$ in the opposite direction then

$m v_A = 2 m v_B$

$\Rightarrow v_A = 2 v_B$

Now we need to compute the potential energy lost which is made up of gravitational PE of $2mgl$ and electrostatic PE. The electrostatic PE is given by the work done, $W$ , to take $B$ from distance $l$ away from the rod to the initial point so

$W = \int_{r}^{l+r} F_E \, \mathrm{d} \rho$

$= \frac{Q \lambda}{2 \pi \epsilon_0 } \ln \left( \frac{l+r}{r} \right)$.

Hence the total KE must be

$\text{KE} = 2mgl + \frac{Q \lambda}{2 \pi \epsilon_0 } \ln \left( \frac{l+r}{r} \right)$

so if the speed of $B$ when the string is vertical is $v$ then we have

$\frac{1}{2} \, (2m) \, v^2 + \frac{1}{2} \, m \, (2v)^2 = 2mgl + \frac{Q \lambda}{2 \pi \epsilon_0 } \ln \left( \frac{l+r}{r} \right)$.

$
\Leftrightarrow 3m v^2 = 2mgl + \frac{Q \lambda}{2 \pi \epsilon_0 } \ln \left( \frac{l+r}{r} \right)$

so

$\color[rgb]{0,0,1} \boxed{v^2 = \frac{2}{3} g l + \frac{Q \lambda}{6 \pi \epsilon_0 m} \ln \left( \frac{l+r}{r} \right)}$.

Finally to find the tension in the string we simply use Newton II to relate the centripetal acceleration to the weight of $B$ and the tension, $T$, so

$2m \frac{v^2}{l} = T - 2mg - F_E$

$\Leftrightarrow T = 2mg + 2m \frac{v^2}{l} + F_E (r+l)$
$= 2mg + \frac{4}{3} m g + \frac{Q \lambda}{3 \pi \epsilon_0 l} \ln \left( \frac{l+r}{r} \right) + \frac{Q \lambda}{2 \pi \epsilon_0 (r+l)}$,

so

$\color[rgb]{0,0,1} \boxed{ T = \frac{10}{3} m g + \frac{Q \lambda}{3 \pi \epsilon_0 l} \ln \left( \frac{l+r}{r} \right) + \frac{Q \lambda}{2 \pi \epsilon_0 (r+l)} }$.

Let me know if I've made any silly algebraic/arithmetic errors.

5. Perfectly in order

My text gives the tension as $T = 2mg + 9\cdot \frac{mv^2}{l} + \left(\frac{\lambda Q}{(2\pi\epsilon_0)\cdot (r + l)}\right)$. I would have probably thought that it is a typo (9 instead of 2 in the $\frac{mv^2}{l}$). But what is that extra term which I cannot make any sense of?

6. Oops! I forgot to include the electrostatic force term when computing the tension in the string. I've corrected it now.