Physics Solution – Question 5

Solution to Question 5

1. Geometry and Symmetry Analysis

We consider three identical point charges $q$ placed at the vertices A, B, and C of an equilateral triangle of side $a$. We are looking for points within the triangle where the net electric field vanishes (null points).

By symmetry, any null point must lie on the axes of symmetry of the system, which are the medians of the triangle.

Figure 1: Distribution of 4 null points within the triangle. One at the centroid and three others located on the medians.

2. Field Analysis along the Median

Let us analyze the electric field along the median AD (where D is the midpoint of BC). Let the distance from the midpoint D upwards towards A be $y$.

Field due to base charges (B and C):
The pair of charges at B and C act like a charged ring. At a point on the bisector at distance $y$ from the midpoint, their fields add up constructively along the median and cancel perpendicular to it.

$$ E_{BC} = \frac{2kq y}{(y^2 + (a/2)^2)^{3/2}} \quad (\text{directed towards A}) $$

At $y=0$, $E_{BC} = 0$. The slope is positive initially, reaching a maximum at $y = \frac{a}{2\sqrt{2}}$.

Field due to top charge (A):
The distance from charge A to the point is $(h-y)$, where $h = \frac{\sqrt{3}}{2}a$ is the altitude.

$$ E_{A} = \frac{kq}{(h-y)^2} \quad (\text{directed away from A, i.e., downwards}) $$

At $y=0$, $E_A = \frac{kq}{h^2}$ which is non-zero.

3. Graphical Intersection and Null Points

A null point occurs when the magnitude of the upward field equals the magnitude of the downward field: $$ E_{BC} = E_A $$

Comparing the two functions near the base ($y \to 0$):

The field $E_{BC}$ starts at 0 and rises sharply with a slope proportional to $\frac{1}{(a/2)^3}$.
The field $E_A$ starts at a finite value $\frac{kq}{h^2}$ but rises more slowly initially as its effective distance parameter is larger ($h > a/2$).

Because the “ring-like” field $E_{BC}$ rises faster initially than the hyperbolic field $E_A$, the curve for $E_{BC}$ overtakes $E_A$ resulting in the first null point located between the midpoint and the centroid.

As we move further up towards the centroid ($y = h/3$):

At the centroid, geometry dictates perfect cancellation due to symmetry ($E_{BC} = E_A$). This is the second null point on this median.

Thus, on the median AD, there are 2 null points.

4. Conclusion

Since the equilateral triangle has 3 identical medians, and the solution structure is symmetric:

There is 1 null point at the common intersection (Centroid).
There is 1 additional null point on each of the 3 medians (located between the centroid and the side).

Total number of null points = $1 (\text{Centroid}) + 3 (\text{on Medians}) = 4$.

Answer: 4 The electric field vanishes at 4 distinct places within the triangle.