Problem: Electric Field Lines & Point Charges
Given: A line of electric field created by $q_1 = 1 \mu\text{C}$ and an unknown charge $q_2$. We need to find $q_2$.
Step 1: The Condition for the Peak
At the highest point of the field line, the tangent is horizontal. This means the net vertical electric field is zero ($E_y = 0$). The upward vertical field from $q_1$ balances the downward vertical field from $q_2$.
$$E_{y1} = E_{y2}$$
Step 2: Geometry from Grid
Using the grid squares (see red lines in diagram):
- Height ($h$) = 4 units
- Horizontal distance from $q_1$ = 2 units
- Horizontal distance from $q_2$ = 8 units
Calculate hypotenuse squared ($r^2$) using Pythagoras:
$$r_1^2 = 2^2 + 4^2 = 20$$
$$r_2^2 = 8^2 + 4^2 = 80$$
Step 3: The Calculation
Using the condition $E_{y1} = E_{y2}$ and substituting $E \propto \frac{q}{r^2}$ and $\sin(\theta) = \frac{h}{r}$:
$$q_2 = q_1 \left( \frac{r_2}{r_1} \right)^3$$
Substituting the values:
$$q_2 = 1 \mu\text{C} \times \left( \frac{80}{20} \right)^{3/2}$$
$$q_2 = 1 \times (4)^{3/2}$$
$$q_2 = 1 \times 8 = 8 \mu\text{C}$$
Step 4: Determine the Sign
Electric field lines originate from positive charges and terminate on negative charges. Since the line leaves $q_1$ and enters $q_2$, $q_2$ must be negative.
Answer: (d) $-8 \mu\text{C}$