Six equal positive charges, q, are kept at six corners of a regular hexagon of sides ‘a’ charge ‘- 6q ‘ is placed at the centre, calculate the potential and hence electric intensity at a large distance r.

To tackle this problem, we need to analyze the electric potential and electric field intensity created by the charges arranged in a hexagon with a negative charge at the center. Let's break this down step by step.

Understanding the Setup

We have six equal positive charges, each of magnitude \( q \), positioned at the corners of a regular hexagon. The distance between adjacent charges is \( a \). At the center of this hexagon, there is a negative charge of magnitude \( -6q \). Our goal is to find the electric potential and the electric field intensity at a large distance \( r \) from the center of the hexagon.

Electric Potential Calculation

The electric potential \( V \) at a point in space due to a point charge is given by the formula:

• \( V = \frac{k \cdot Q}{r} \)

where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge to the point where the potential is being calculated.

At a large distance \( r \) from the center of the hexagon, we can treat the system as a collection of point charges. The potential due to each of the six positive charges at a distance \( r \) is:

• \( V_{+} = \frac{k \cdot q}{r} \)

Since there are six such charges, the total potential due to the positive charges is:

• \( V_{total, +} = 6 \cdot \frac{k \cdot q}{r} = \frac{6kq}{r} \)

Now, considering the negative charge at the center, the potential due to this charge at a distance \( r \) is:

• \( V_{-} = \frac{k \cdot (-6q)}{r} = -\frac{6kq}{r} \)

Combining both contributions, the total electric potential \( V_{total} \) at a distance \( r \) from the center is:

• \( V_{total} = \frac{6kq}{r} - \frac{6kq}{r} = 0 \)

Electric Field Intensity Calculation

The electric field intensity \( E \) is related to the electric potential by the gradient of the potential. However, since we have found that the total potential \( V_{total} \) is zero, we can also analyze the contributions to the electric field from each charge.

The electric field due to a point charge is given by:

• \( E = \frac{k \cdot Q}{r^2} \)

For the six positive charges, at a large distance \( r \), the electric field contribution from each charge will point away from the charge. The total electric field due to the six positive charges can be calculated as:

• \( E_{+} = 6 \cdot \frac{k \cdot q}{r^2} \)

For the negative charge at the center, the electric field points towards the charge, and its contribution is:

• \( E_{-} = \frac{k \cdot (-6q)}{r^2} \)

When we sum these contributions, we find:

• \( E_{total} = 6 \cdot \frac{k \cdot q}{r^2} - \frac{6kq}{r^2} = 0 \)

Final Insights

At a large distance \( r \) from the hexagon, both the electric potential and the electric field intensity are zero. This outcome is due to the symmetrical arrangement of the positive charges and the negative charge at the center, which perfectly cancels out their effects at that distance. This scenario illustrates the principle of superposition in electrostatics, where the contributions from multiple charges can combine to yield a net effect of zero under certain conditions.