Bouncing off a charged sheet

The two forces acting on the point charge are the electrical force $F_E = qE$ and the gravitational force $F_G = mg$ . The electric field around a charged sheet (presumed to be sufficiently large) is the same at every point and equal to $E= \frac{\sigma}{2 \varepsilon_0}$ pointing away from the sheet.

Let $\theta$ be the angle that the rod makes with the horizontal, so that $\theta = 90°$ when it is horizontal and $\theta = 0°$ when it points straight down. Then the total torque $\tau$ on the rod is: $\tau = F_El\cos\theta - F_Gl\sin\theta =l \left(\frac{q\sigma}{2\varepsilon_0}\cos\theta - mg\sin\theta \right)$

The moment of inertia of a point mass at a distance $l$ from the axis of rotation is given by $I=ml^2$ . Thus, the angular acceleration $\alpha$ of the charge is:

$\alpha = \frac{\tau}{I} = \frac{1}{ml} \left(\frac{q\sigma}{2\varepsilon_0}\cos\theta - mg\sin\theta \right)$

Since $\omega = vl$ , the angular velocity of the charge is maximum whenever the speed is maximum. This is the case when $\theta = \theta^* = 30°$ . Furthermore, whenever $\omega$ is maximum, it's derivative $\alpha = \frac{\mathrm{d}\omega}{\mathrm{d}t}$ must equal zero. Therefore, we set $\alpha$ equal to 0 and solve for $\sigma$ : $\alpha = \frac{1}{ml} \left(\frac{q\sigma}{2\varepsilon_0}\cos\theta^* - mg\sin\theta^* \right) = 0$ $\Leftrightarrow \frac{q\sigma}{2\varepsilon_0}\cos\theta^* - mg\sin\theta^* = 0$ $\Leftrightarrow \frac{q\sigma}{2\varepsilon_0} = mg\tan\theta^* \Leftrightarrow \sigma = \frac{2\varepsilon_0mg}{q}\tan\theta^*$

Inserting the given values, we get: $\sigma = 1.00 \times 10^{-8}\mathrm{C/m^2}$

In order for its speed to be maximum, the acceleration of the particle (and therefore the net force acting on the particle) must be $0$ . Therefore, when the particle makes an angle of $30^\circ$ , the net force acting on it is $0$ . In this problem, the only forces that can act on the particle are those from gravity ( $F_g$ , only in the negative z-direction), the rod ( $F_r$ , in the positive x- and positive z-direction), and the charged sheet ( $F_e$ , only in the negative x-direction).

We know that $F_g = m\cdot{g} = .1[kg]\cdot9.8[\frac{m}{s^2}] = .98[N]$ in the negative z-direction and that the component of $F_r$ that acts in the positive z-direction, $F_{rz}$ , must be equal to $F_g$ in magnitude in order for the net force in the z-direction to be $0$ . Therefore, $F_{rz} = .98[N]$ . From this and the fact that the rod makes a $30^\circ$ angle with the vertical, we find that $F_{rx}$ , the x-component of $F_r$ , is $.98[N]\cdot{\cot(30^\circ)} = \frac{.98}{\sqrt{3}}[N]$ in the positive x-direction, and so $F_e = \frac{.98}{\sqrt{3}}[N]$ in the negative x-direction because the net force acting on the particle in the x-direction must be $0$ .

To find the electrostatic force on a point of charge $q[C]$ from an electric field with strength (magnitude) $E[\frac{N}{C}]$ , we must multiply $E$ by $q$ . Thus, the strength of the electric field at the point charge from the charged sheet is $\frac{\frac{.98}{\sqrt{3}}[N]}{.001[C]} = \frac{980}{\sqrt{3}}[\frac{N}{C}]$ . Using Gauss's Law, we can derive that the strength of the electric field given off by a charged sheet with uniform charge density is dependent only on $\sigma$ , its charge density. In fact, the strength is given by $E = \frac{\sigma}{2\epsilon_0}[\frac{N}{C}]$ . We then have that $\frac{\sigma}{2\epsilon_0}[\frac{N}{C}] = \frac{980}{\sqrt{3}}[\frac{N}{C}]$ , and solving this equation for $\sigma$ gives us $\sigma = \frac{980\cdot{2\epsilon_0}}{\sqrt{3}}[\frac{C}{m^2}] =$ $1.00056\cdot10^{-8}[\frac{C}{m^2}]$ . Rounding to $3$ significant digits gives us a final answer of $\sigma = 1.00\cdot10^{-8}[\frac{C}{m^2}]$ .

At 30\degree the point charge is at its maximum speed. Thus this means the point charge has stopped accelerating. Thus the tension in the string pulling back equals the electrostatic force. In addition the tension pulling up must equal the gravitational force. So we can now state that tan(30)=\frac {electrostatic force}{gravitational force}. The gravitational force is just mg. To get the electrostatic force, we know that F=qE. To get E, we use Gauss's Law. So \oint E\dot dA=\frac {Q}{\epsilon}. We just a cylinder as our Gaussian and only the bases will have electric flux going through it. So \oint E\dot dA = E2A=\frac {Q}{\epsilon}. \sigma =\frac {Q}{A}. So Q=A \dot \sigma. Combing these two equations, we get E2A=\frac {A \dot \sigma}{\epsilon} \rightarrow E=\frac {\sigma}{2 \epsilon}. Finally putting this and the gravitational force into our first equation, we get tan(30)=\frac {q \sigma}{2 \epsilon mg}. Putting in our values for q and m, we get \sigma=1E-8.

The maximum speed will occur when the potential energy is a minimum (since all forces are conservative). We first assign the zero of potential when the point charge is at the lowest height possible, i.e. the rod is parallel to the sheet, straight down from where it's attached. The potential energy is therefore $U=mgL(1-\cos \theta) - \frac{L\sigma q}{2 \epsilon_0} \sin \theta$ where $\theta$ is the angle with respect to the vertical. We find the minimum by taking the derivative and setting it equal to zero, which yields

$\sigma=\frac{2\epsilon_0 mg \tan \theta}{q}$

Substituting in numbers gives $\sigma=1 \times 10^{-8}~\mbox{C/m}^2$ .

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σ =(0.1 9.8 2)/(1.73 9 4 3.14 0.000000000001) σ =1.001* 10^(-8)