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: \( q_1 = q_2 = q_3 = 1 \, \text{pC} = 1 \times 10^{-12} \, \text{C} \)
Equilateral triangle side: \( a = 10 \, \text{cm} = 0.1 \, \text{m} \)

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

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

Electric fields from vertex charges:
1. Field from \( q_1 \): \[ E_1 = \frac{9 \times 10^9 \cdot 1 \times 10^{-12}} {\left(\frac{0.1}{\sqrt{3}}\right)^2} = 2.7 \, \text{N/C} \] Direction: centroid → \( q_1 \).

2. Field from \( q_2 \): \[ E_2 = 2.7 \, \text{N/C} \] Direction: centroid → \( q_2 \).

3. Field from \( q_3 \): \[ E_3 = 2.7 \, \text{N/C} \] Direction: centroid → \( q_3 \).

Step 3: Resolve electric fields into components.
Due to symmetry, \( E_1 \), \( E_2 \), and \( E_3 \) are \( 120^\circ \) apart. Their components cancel out.

Step 4: Determine the net electric field.
The symmetrically distributed electric fields cancel each other. Therefore, the net electric field at the centroid is zero.

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