ABC is an equilateral triangle. Charges +q are placed at each corner. The electric field intensity at the centroid of triangle will be

Options

A. `(1)/(4pi epsilon_(0))xx(q)/(r^(2))`
B. `(1)/(4pi epsilon_(0))xx(3q)/(r^(2))`
C. `(1)/(4pi epsilon_(0))xx(q)/( r )`
D. Zero

Answer

To find the electric field intensity at the centroid of an equilateral triangle with charges \(+q\) placed at each corner, follow these steps:

### Step-by-Step Solution:

1. **Identify the Setup:**
- Consider an equilateral triangle \(ABC\) with side length \(a\).
- Charges \(+q\) are placed at each corner \(A\), \(B\), and \(C\).
- We need to find the electric field intensity at the centroid \(O\) of the triangle.

2. **Electric Field Due to a Single Charge:**
- The electric field due to a single charge \(+q\) at a distance \(r\) is given by:
\[ E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2} \]
- In an equilateral triangle, the distance from each vertex to the centroid \(O\) is:
\[ r = \frac{a}{\sqrt{3}} \]

3. **Calculate the Electric Field Magnitude:**
- The electric field due to charge \(+q\) at one vertex (say \(A\)) at the centroid \(O\) is:
\[ E_A = \frac{1}{4 \pi \epsilon_0} \frac{q}{\left(\frac{a}{\sqrt{3}}\right)^2} = \frac{1}{4 \pi \epsilon_0} \frac{q \cdot 3}{a^2} = \frac{3q}{4 \pi \epsilon_0 a^2} \]

4. **Direction of Electric Fields:**
- The electric field vectors due to charges at \(A\), \(B\), and \(C\) will point away from each charge because they are positive charges.
- The angle between any two electric field vectors at the centroid is \(120^\circ\).

5. **Resultant Electric Field Calculation:**
- Let \(E_A\), \(E_B\), and \(E_C\) be the electric fields due to charges at \(A\), \(B\), and \(C\) respectively.
- Since \(E_A = E_B = E_C\) in magnitude, we can use vector addition to find the resultant electric field.
- The resultant electric field \(E_R\) at the centroid \(O\) due to two charges (say \(A\) and \(B\)) is given by:
\[ E_{AB} = \sqrt{E_A^2 + E_B^2 + 2E_AE_B \cos(120^\circ)} \]
- Since \(\cos(120^\circ) = -\frac{1}{2}\):
\[ E_{AB} = \sqrt{E_A^2 + E_A^2 + 2E_AE_A \left(-\frac{1}{2}\right)} = \sqrt{E_A^2 + E_A^2 - E_A^2} = E_A \]
- Similarly, the resultant of \(E_{AB}\) and \(E_C\) will also be zero because they are equal in magnitude and symmetrically opposite.

6. **Conclusion:**
- The electric fields due to the three charges at the centroid \(O\) cancel each other out.
- Therefore, the net electric field intensity at the centroid \(O\) is:
\[ E_O = 0 \]

### Final Answer:
The electric field intensity at the centroid of the triangle is \(0\).