Problem 43 - A Complete Resource Book in Physics for JEE Main

Question

In a regular polygon of \(n\) sides, each corner is at a distance \(r\) from the centre. Identical charges of magnitude \(Q\) are placed at \((n-1)\) corners. The field at the centre is (A) \(k \frac{Q}{r^{2}}\) (B) \((n-1) k \frac{Q}{r^{2}}\) (C) \(\frac{n}{n-1} k \frac{Q}{r^{2}}\) (D) \(\frac{n-1}{n} k \frac{Q}{r^{2}}\)

Step-by-Step Solution

Answer

The short answer is: The electric field at the center of the regular polygon is given by \((n-1) k \frac{Q}{r^{2}}\).

Step 1: Determine the angle between the charges

Since the polygon is regular, all angles between the sides will be equal. The sum of angles in an n-sided polygon is \((n-2) \times 180^{\circ}\). Therefore, each angle will be \(\frac{(n-2) \times 180^{\circ}}{n}\).

Step 2: Calculate the angle bisector

Since the field at the center is due to all the charges placed at the corners, we need to find the angle bisector for each charge, which will be half of the angle between the charges. Let's call this angle \(\theta\), which will be equal to \(\frac{(n-2) \times 180^{\circ}}{2n}\).

Step 3: Calculate the electric field due to one charge

We can calculate the electric field due to one charge using the formula \(E = k \frac{q}{r^{2}}\), substituting Q for q and r for r: \(E_1 = k \frac{Q}{r^{2}}\)

Step 4: Calculate the total electric field at the center

To find the total electric field at the center, we need to account for the components of the electric fields in the horizontal and vertical direction. The horizontal component of the electric field due to one charge is given by \(E_1 \cos{\theta}\), and the vertical component is given by \(E_1 \sin{\theta}\). For the total electric field in the horizontal direction, we must sum up the horizontal components due to all (n-1) charges. Similarly, we must also sum up the vertical components due to all (n-1) charges. Since the polygon is symmetric, the horizontal and vertical components will cancel out, and the magnitude of the electric field will be simply multiplied by (n-1) charges. Hence, the total electric field at the center will be \(E = (n-1)E_1 = (n-1)k\frac{Q}{r^2}\). So, the correct option is: (B) \((n-1) k \frac{Q}{r^{2}}\)