Six equal point charges `q_(0)` each are placed at six corners of a regular hexagon of side 'a'. Find out work required o take a point charge 'q' slowly :
(i) From infinity to the centre of haxagon.
(ii) From infinity to a point on the axis which is at a distance `'sqrt(3) a'` from the centre of hexagon.
(iii) Does your answer to part (i) and (ii) depends on the path followed by the charge.

Answer

To solve the problem step by step, we will analyze each part of the question systematically.

### Given:
- Six equal point charges \( q_0 \) are placed at the corners of a regular hexagon with side length \( a \).
- We need to find the work done to bring a point charge \( q \) from infinity to two different locations: the center of the hexagon and a point on the axis at a distance \( \sqrt{3}a \) from the center.

### Part (i): Work done to bring charge \( q \) to the center of the hexagon

1. **Determine the potential at the center of the hexagon**:
- The potential \( V \) at a point due to a point charge is given by:
\[ V = k \frac{Q}{r} \]
- For six charges \( q_0 \) at the corners of the hexagon, the distance from each charge to the center (O) is \( \frac{a}{\sqrt{3}} \) (using geometry of the hexagon).
- Therefore, the total potential \( V_O \) at the center is:
\[ V_O = k \left( \frac{q_0}{\frac{a}{\sqrt{3}}} + \frac{q_0}{\frac{a}{\sqrt{3}}} + \frac{q_0}{\frac{a}{\sqrt{3}}} + \frac{q_0}{\frac{a}{\sqrt{3}}} + \frac{q_0}{\frac{a}{\sqrt{3}}} + \frac{q_0}{\frac{a}{\sqrt{3}}} \right) = 6k \frac{q_0 \sqrt{3}}{a} \]

2. **Calculate work done**:
- The work done \( W \) to bring the charge \( q \) from infinity (where potential \( V_{\infty} = 0 \)) to the center is given by:
\[ W = q \left( V_O - V_{\infty} \right) = q \left( 6k \frac{q_0 \sqrt{3}}{a} - 0 \right) = 6k \frac{q q_0 \sqrt{3}}{a} \]

### Part (ii): Work done to bring charge \( q \) to a point on the axis at distance \( \sqrt{3}a \)

1. **Determine the potential at the point on the axis**:
- The distance from the center of the hexagon to this point is \( \sqrt{3}a \).
- The potential \( V_P \) at this point due to the six charges is:
\[ V_P = k \left( \frac{q_0}{\sqrt{3}a} + \frac{q_0}{\sqrt{3}a} + \frac{q_0}{\sqrt{3}a} + \frac{q_0}{\sqrt{3}a} + \frac{q_0}{\sqrt{3}a} + \frac{q_0}{\sqrt{3}a} \right) = 6k \frac{q_0}{\sqrt{3}a} \]

2. **Calculate work done**:
- The work done \( W \) to bring the charge \( q \) from infinity to this point is:
\[ W = q \left( V_P - V_{\infty} \right) = q \left( 6k \frac{q_0}{\sqrt{3}a} - 0 \right) = 6k \frac{q q_0}{\sqrt{3}a} \]

### Part (iii): Dependence of work done on the path

- The work done in moving a charge in an electric field depends only on the initial and final positions, not on the path taken. This is because the electric field is conservative.
- Therefore, the answer to part (i) and part (ii) does not depend on the path followed by the charge.

### Summary of Results:
1. Work done to bring charge \( q \) to the center:
\[ W_1 = 6k \frac{q q_0 \sqrt{3}}{a} \]
2. Work done to bring charge \( q \) to the point on the axis:
\[ W_2 = 6k \frac{q q_0}{\sqrt{3}a} \]
3. The work done does not depend on the path followed.