Question

4 Points charges each \(+q\) is placed on the circumference of a circle of diameter \(2 \mathrm{~d}\) in such a way that they form a square. The potential at the centre is \(\ldots \ldots .\) (A) 0 (B) \((4 \mathrm{kd} / \mathrm{q})\) (C) \((\mathrm{kd} / 4 \mathrm{q})\) (D) \((4 \mathrm{kq} / \mathrm{d})\)

Short answer

The electric potential at the center of the circle is \(\frac{4kq}{d}\), which corresponds to option (D).

Step-by-step solution

Calculate side length of the square

Since the square is inscribed in a circle with diameter \(2d\), the diagonal of the square = diameter of the circle. We'll use the Pythagorean theorem to find the side length of the square: \(s^2 + s^2 = (2d)^2\) Solving for \(s\), we get: \(s = d\sqrt{2}\)

Calculate distance of each charge to the center

We'll use the Pythagorean theorem again to find the distance from each corner of the square (where the charges are placed) to the center of the circle. As the charges are equidistant from the center, we can find the distance of any one charge to the center. \(\frac{s}{2}^2 + \frac{s}{2}^2 = r^2\) Substituting the value of \(s\), we get: \(\frac{d\sqrt{2}}{2}^2 + \frac{d\sqrt{2}}{2}^2 = r^2\) On solving for \(r\), we get \(r = d\) #Step 2: Calculate the electric potential at the center of the circle#

Calculate the potential due to each charge

Using the formula for electric potential due to a point charge (\(V = k\frac{q}{r}\)), we can calculate the potential at the center due to one point charge: \(V_{1charge} = k\frac{q}{d}\)

Calculate the total potential at the center

Since there are 4 point charges and electric potential is a scalar quantity, we simply sum the contributions from each charge: \(V_{total} = 4 \times V_{1charge} = 4 \times k\frac{q}{d} = \frac{4kq}{d}\) So, the electric potential at the center of the circle is \(\frac{4kq}{d}\), which matches option (D).