ELECTROSTATICS O30-32

Process: For $0 \to 0.5 T_0$, Jockey is at A (Force UP). For $0.5 T_0 \to T_0$, Jockey is at B (Force DOWN).

Phase 1 ($t = 0$ to $t_1 = T_0/2$):

$$ v_{y1} = a_0 t_1 = a_0 \frac{T_0}{2} $$ $$ y_1 = \frac{1}{2} a_0 t_1^2 = \frac{1}{2} a_0 \left(\frac{T_0}{2}\right)^2 = \frac{a_0 T_0^2}{8} $$

Phase 2 ($t = T_0/2$ to $T_0$):

Acceleration reverses: $a = -a_0$. Duration $\Delta t = T_0/2$.

$$ y_{final} = y_1 + v_{y1} \Delta t + \frac{1}{2} (-a_0) (\Delta t)^2 $$ $$ y_{final} = \frac{a_0 T_0^2}{8} + \left(a_0 \frac{T_0}{2}\right)\frac{T_0}{2} – \frac{1}{2} a_0 \frac{T_0^2}{4} $$ $$ y_{final} = \frac{a_0 T_0^2}{8} + \frac{a_0 T_0^2}{4} – \frac{a_0 T_0^2}{8} = \frac{a_0 T_0^2}{4} $$

Substitute $a_0 = \frac{eV}{2md}$:

$$ y_{final} = \frac{1}{4} \left( \frac{eV}{2md} \right) T_0^2 = \frac{eV T_0^2}{8md} $$

Since $y > 0$, it is above the x-axis.

Process: Jockey slides C $\to$ A $\to$ C. This creates a triangular voltage pulse. The force varies linearly with time.

Force $F_y \propto t$. Acceleration $a_y(t)$ forms a triangle graph (0 to $a_{max}$ to 0) over time $T_0$.

Velocity Calculation:

$$ v_y = \int_0^{T_0} a_y(t) dt = \text{Area under a-t graph} $$ $$ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} T_0 \times a_{max} $$

Where $a_{max}$ occurs at A: $a_{max} = \frac{eV}{2md}$.

$$ v_y = \frac{1}{2} T_0 \left( \frac{eV}{2md} \right) = \frac{eV T_0}{4md} $$

Horizontal velocity $v_x = l/T_0$. Total velocity vector:

$$ \vec{v} = \frac{l}{T_0} \hat{i} + \frac{eV T_0}{4md} \hat{j} $$

Since the force was always positive (C to A is positive potential), the displacement is positive (above axis).

Process: Jockey slides A $\to$ B with constant speed. Voltage goes linearly from $+V/2$ to $-V/2$. Acceleration goes linearly from $+a_0$ to $-a_0$.

Velocity Calculation:

$$ v_y = \int_0^{T_0} a_y(t) dt $$

The integral of a linear function going from $+a_0$ to $-a_0$ symmetrically over time $T_0$ is zero (Positive area cancels negative area).

$$ v_y(\text{final}) = 0 $$

Thus, the final velocity vector is just horizontal: $\vec{v} = \frac{l}{T_0} \hat{i}$.

Position Calculation:

Since $v_y$ starts at 0, increases, and returns to 0 (always positive during the interval), the displacement $y$ will be positive.

Calculation: $y = \int v_y dt$. Since $v_y(t)$ is a parabola opening downwards (integral of linear with negative slope) and is positive, the area is positive.