Question

Three point charges, 1 pC each, are kept at the vertices of an equilateral triangle of side 10 cm. Find the net electric field at the centroid of the triangle.

Hint:

For symmetrically placed charges in an equilateral triangle, the resultant electric field at the centroid can be found by calculating the field due to one charge and using the symmetry to sum the components.

Answer 1

Given: Charges at the vertices: \( q_1 = q_2 = q_3 = 1 \, \text{pC} = 1 \times 10^{-12} \, \text{C} \)
Side of the equilateral triangle: \( a = 10 \, \text{cm} = 0.1 \, \text{m} \)

Step 1: Calculate the distance from the centroid to a vertex.
For an equilateral triangle, the distance (\( r \)) from the centroid to any vertex is: \[ r = \frac{a}{\sqrt{3}} = \frac{0.1}{\sqrt{3}} \, \text{m} \]

Step 2: Calculate the electric field due to each charge.
The electric field (\( E \)) due to a point charge is given by: \[ E = \frac{k \cdot q}{r^2} \] where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

Electric fields due to charges at vertices:
1. **Electric field due to \( q_1 \):** \[ E_1 = \frac{9 \times 10^9 \cdot 1 \times 10^{-12}} {\left(\frac{0.1}{\sqrt{3}}\right)^2} = 2.7 \, \text{N/C} \] Direction of \( E_1 \): along centroid → \( q_1 \).

2. **Electric field due to \( q_2 \):** \[ E_2 = 2.7 \, \text{N/C} \] Direction: along centroid → \( q_2 \).

3. **Electric field due to \( q_3 \):** \[ E_3 = 2.7 \, \text{N/C} \] Direction: along centroid → \( q_3 \).

Step 3: Resolve the electric fields into components.
Due to symmetry, \( E_1 \), \( E_2 \), and \( E_3 \) are at \( 120^\circ \) to each other. Their horizontal and vertical components cancel out.

Step 4: Calculate the net electric field.
Since the electric fields are symmetrically distributed and cancel out, the net electric field at the centroid is zero.

Final Answer: \[ \boxed{\text{The net electric field at the centroid is } 0 \, \text{N/C.}} \]

Answer 2

Step 1: Geometry — distance from centroid to each vertex

For an equilateral triangle, the distance \(R\) from the centroid (which coincides with the circumcenter) to any vertex is \[ R = \frac{a}{\sqrt{3}} = \frac{0.1}{\sqrt{3}} \,\text{m}. \]

Step 2: Field due to one charge at the centroid

Magnitude from a single charge \(q=1\,\text{pC}=10^{-12}\,\text{C}\): \[ E_1 = \frac{k q}{R^2} = \frac{k \cdot 10^{-12}}{\left(\frac{a}{\sqrt{3}}\right)^2} = \frac{k \cdot 10^{-12}}{a^2/3} = \frac{3k \cdot 10^{-12}}{a^2}. \] With \(k=\frac{1}{4\pi\varepsilon_0}\approx 8.99\times 10^9\,\text{N m}^2\!\!/\text{C}^2\) and \(a=0.1\,\text{m}\), this gives \(E_1 \approx 2.7\,\text{N/C}\) (direction: along the line from the vertex toward the centroid).

Step 3: Vector sum by symmetry

The three charges are identical and symmetrically placed, so the three field vectors at the centroid have the same magnitude \(E_1\) and are separated by \(120^\circ\). The vector sum of three equal vectors spaced by \(120^\circ\) is zero: \[ \vec E_{\text{net}} = \vec E_1 + \vec E_2 + \vec E_3 = E_1\!\left(1 + e^{i2\pi/3} + e^{i4\pi/3}\right) = \vec 0. \] (Geometrically, they close a triangle of vectors.)

Net electric field at the centroid: \[ \boxed{\vec E_{\text{net}} = \vec 0 \;\; (\text{zero})}. \]

This result holds for any three equal charges at the vertices of an equilateral triangle, regardless of the side length—by symmetry the field at the center cancels out.