Solution
The problem involves a charged particle in equilibrium under the influence of gravity and an electric force. The equilibrium condition implies:
$$mg = qE$$Thus, the required charge is inversely proportional to the electric field strength: $q \propto \frac{1}{E}$.
We need to determine the ratio of the electric field strength at the two equilibrium positions: position 1 $(0, 10)$ and position 2 $(0, 0)$. We can use Gauss’s Law for a flux tube. The problem states the field is symmetrical about the y-axis.
Consider a circular flux tube centered on the y-axis. The electric flux $\Phi_E$ through any cross-section of this tube is constant:
$$\Phi_E = E \cdot A = E \cdot (\pi r^2) = \text{constant}$$ $$E \propto \frac{1}{r^2}$$From the grid in the diagram, we can trace a specific field line to find the ratio of radii:
- At $y = 0$ (bottom), let’s look at the field line starting at $x \approx 5$ units. So, $r_1 = 5$.
- Follow this line to $y = 10$. It reaches $x \approx 15$ units. So, $r_2 = 15$.
Comparing the fields at these two heights:
$$\frac{E(y=0)}{E(y=10)} = \frac{r_2^2}{r_1^2} = \left(\frac{15}{5}\right)^2 = 3^2 = 9$$ $$E(y=0) = 9 E(y=10)$$The field at the bottom $(0,0)$ is 9 times stronger than at $(0,10)$.
Applying the equilibrium condition:
- Case 1 at $(0,10)$: $q_{old} E_{10} = mg$
- Case 2 at $(0,0)$: $q_{new} E_{0} = mg$
The change in charge is $\Delta q = |q_{new} – q_{old}| = | \frac{q}{9} – q | = \frac{8}{9}q$.
The fractional change is:
$$\frac{\Delta q}{q} = \frac{8}{9}$$Answer: (d) 8/9