Problem 3: Analysis of Electric Field vs. Time
1. Modulus of the Charges
At $t \to \infty$, the moving charge is infinitely far away. The electric field $E_0$ is generated solely by the fixed charge located at distance $r_0$ from the observation point P.
$$ E_0 = \frac{1}{4\pi\epsilon_0} \frac{q}{r_0^2} $$Rearranging for charge $q$:
2. Minimum Distance from P
At $t = \tau$, the net electric field is zero. This implies the fields from the two identical charges cancel out perfectly. This can only happen if they are on opposite sides of P at equal distances.
Since the fixed charge is at distance $r_0$, the moving charge must also be at distance $r_0$ from P at this instant. Due to the symmetry of the graph (even function around $\tau$), the path must be perpendicular to the line joining the fixed charge and P.
3. Speed of the Moving Charge
We analyze the slope of the $E-t$ graph near the zero crossing ($t=\tau$). Let the moving charge move along the perpendicular path with velocity $v$. Its distance from the axis is $y = v(t-\tau)$.
For small $y$, the net field is dominated by the perpendicular component of the moving charge’s field:
$$ E_{net} \approx E_y = E_{moving} \sin(\theta) \approx \frac{kq}{r_0^2} \frac{y}{r_0} = E_0 \frac{v(t-\tau)}{r_0} $$The slope of the field with respect to time is:
$$ \frac{dE}{dt} = \frac{E_0 v}{r_0} $$Looking at the provided graph, the tangent at the zero crossing intersects the level $E_0$ at time $t = \tau + t_0$. Therefore, the geometric slope is:
$$ \text{Slope} = \frac{E_0}{t_0} $$Equating the physical slope to the geometric slope:
$$ \frac{E_0 v}{r_0} = \frac{E_0}{t_0} \implies v = \frac{r_0}{t_0} $$