Three equal charges Q are placed at the three corners of an equilateral triangle of side length a. Work done in shifting a charge q from infinity to the centroid of the triangle is

Options

A. `(3Qq)/(4pi in_(0)a)`
B. `(3Qq)/(4pi in_(0)sqrt3a)`
C. `(3sqrt3Qq)/(4pi in_(0)a)`
D. `(sqrt3Qq)/(4pi in_(0)3a)`

Explanation

To solve the problem of finding the work done in shifting a charge \( q \) from infinity to the centroid of an equilateral triangle with three equal charges \( Q \) at its corners, we can follow these steps:

### Step 1: Understand the Configuration
We have an equilateral triangle with side length \( a \) and three equal charges \( Q \) placed at each corner. The centroid of the triangle is the point where the three medians intersect, and it is equidistant from all three corners.

### Step 2: Determine the Distance from the Centroid to Each Charge
For an equilateral triangle, the distance \( D \) from the centroid to any vertex (corner) can be calculated using the formula:
\[ D = \frac{a}{\sqrt{3}} \]
This is derived from the geometry of the triangle.

### Step 3: Calculate the Electric Potential at the Centroid
The electric potential \( V \) at the centroid due to one charge \( Q \) is given by:
\[ V = \frac{kQ}{D} \]
where \( k = \frac{1}{4\pi \epsilon_0} \) is the Coulomb's constant.

Since there are three charges, the total potential \( V_{\text{total}} \) at the centroid is:
\[ V_{\text{total}} = 3 \cdot \frac{kQ}{D} = 3 \cdot \frac{kQ}{\frac{a}{\sqrt{3}}} = \frac{3\sqrt{3}kQ}{a} \]

### Step 4: Calculate the Work Done in Bringing Charge \( q \)
The work done \( W \) in bringing a charge \( q \) from infinity (where the potential is zero) to the centroid (where the potential is \( V_{\text{total}} \)) is given by:
\[ W = q \cdot V_{\text{total}} = q \cdot \frac{3\sqrt{3}kQ}{a} \]

### Step 5: Substitute the Value of \( k \)
Substituting \( k = \frac{1}{4\pi \epsilon_0} \) into the equation gives:
\[ W = q \cdot \frac{3\sqrt{3} \cdot \frac{1}{4\pi \epsilon_0} Q}{a} = \frac{3\sqrt{3}qQ}{4\pi \epsilon_0 a} \]

### Final Answer
Thus, the work done in shifting the charge \( q \) from infinity to the centroid of the triangle is:
\[ W = \frac{3\sqrt{3}qQ}{4\pi \epsilon_0 a} \]

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